Eamcet Mathematics - Circle Chapter Sample Question Paper With Answer Key
Exam Duration: 60 Mins Total Questions : 50
Length of the diameter of circle \({ X }^{ 2 }+{ Y }^{ 2 }-8X+6Y-5=0\) is
- (a)
\(4\sqrt{30}\)
- (b)
\(2\sqrt{30}\)
- (c)
\(3\sqrt{30}\)
- (d)
30
Radius = \(\sqrt { { \left( -4 \right) }^{ 2 }+{ \left( 3 \right) }^{ 2 }-\left( -5 \right) } =\sqrt { 16+9+5 } =\sqrt { 30 } \)
\(\therefore \) Diameter = \(2\sqrt{30}\)
Equation of the circle with centre on the Y-axis and passing through the origin and the point (2, 3) is
- (a)
\({ x }^{ 2 }+{ y }^{ 2 }+13y=0\)
- (b)
\({ 3x }^{ 2 }+{ 3y }^{ 2 }+13x+3=0\)
- (c)
\({ 6x }^{ 2 }+{ 6y }^{ 2 }-26y=0\)
- (d)
\({ x }^{ 2 }+{ y }^{ 2 }+13x+3=0\)
Let the centre of circle be (0, k).
Let equation of circle be
x2+y2+2(0)x+2ky+c=0
⇒ x2+y2-2ky+c=0
Since, it passes through (O, O) and (2, 3). and 13-6k+c=0
\(⇒13-6k+0=0⇒k={13\over 6}\)
On putting the values of k and c in Eq(i), we get
\(x^2+y^2-2\left(13\over6\right)y+0=0\)
⇒ 6x2+6y2-26y=0
Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line Y - 4X + 3 = 0.
- (a)
\({ x }^{ 2 }+{ y }^{ 2 }-4x-10y+25=0\)
- (b)
\({ x }^{ 2 }+{ y }^{ 2 }-4x-10y-25=0\)
- (c)
\({ x }^{ 2 }+{ y }^{ 2 }-4x+10y-25=0\)
- (d)
None of the above
Let the equation of circle be
x2+y2+2gx+2fy+c=0 ...(i)
Since, it passes through the points (2, 3) and (4, 5).
∴ 22+22+2g(2)+2f(3)+c=0
⇒ 4g+6f+c+13=0 ...(ii)
and 42 + 52 + 2g(2)+2f(5)
⇒ 8g+10f+C+ 41=0
On subtracting Eq. (ii) from Eq. (iii), we get
4g + 4f 28-0
⇒ g+f+7=0
Also, centre(-g, -f) lies on line y - 4x + 3=0.
∴ -f+4g+3=0
On solving Eqs. (iv) and (v), we get
⇒ g=-2, f=-5
On putting g = -2, f=-5 in Eq. (ii), we get
4(-2)+6(-5)+c+13=0⇒c=25= 25
On putting values of g, fand c in Eq. (i), we get
x2 + y2-4x-10y+25=0
\({ X }^{ 2 }+{ Y }^{ 2 }+4X-6Y+4=0\) is the equation of the incircle of an equilateral triangle, then the equation of the circumcircle of the triangle is
- (a)
\({ x }^{ 2 }+{ y }^{ 2 }+4x+6y-23=0\)
- (b)
\({ x }^{ 2 }+{ y }^{ 2 }+4x-6y-23=0\)
- (c)
\({ x }^{ 2 }+{ y }^{ 2 }-4x-6y-23=0\)
- (d)
None of the above
The given incircle is \({ x }^{ 2 }+{ y }^{ 2 }+4x-6y-4=0\)
\(\therefore \) Incentre is (- 2, 3) and inradius = \(\sqrt { 4+9-4 } =3\)
Since, in an equilateral triangle, the incentre and the circumcentre coincide.
\(\therefore \) Circumcentre \(\equiv \) ( - 2, 3)
Also, in an equilateral triangle, circumradius = 2 (inradius)
\(\therefore \) Circumradius = 2 X 3 = 6
\(\therefore \) The equation of the circumcircle is \({ \left( X+2 \right) }^{ 2 }+{ \left( Y-3 \right) }^{ 2 }={ \left( 6 \right) }^{ 2 }\)
\(\Rightarrow \) \({ x }^{ 2 }+{ y }^{ 2 }+4x-6y-23=0\)
The circle \({ X }^{ 2 }+{ Y }^{ 2 }-2X+4Y+3=0\) has a square inscribed in it. Sides of the square are parallel to the coordinate axes. Then, one vertex of the square is
- (a)
\(\left( 1+\sqrt { 2 } ,-2 \right) \)
- (b)
\(\left( 1-\sqrt { 2 } ,-2 \right) \)
- (c)
\(\left( 1,-2+\sqrt { 2 } \right) \)
- (d)
None of the above
Clearly, (1, -2) is the centre of the given circle. Since, the sides of the square, inscribed in the circle, are parallel to the coordinate axes, therefore coordinate of any vertex cannot be equal to - 2 and its X-coordinate cannot be equal to 1. Hence, none of the points given in (a), (b) and (c) can be the vertex of the square.
If the line 3X - 4Y - k = 0, touches the circle \({ X }^{ 2 }+{ Y }^{ 2 }-4X-8Y-5=0\) at (a,b), then k + a + b is equal to
- (a)
20
- (b)
22
- (c)
- 30
- (d)
- 28
Since, the given line touches the given circle, the length of the perpendicular from the centre (2, 4) of the circle to the line 3x-4y-k=0 is equal to the radius \(\sqrt{4+16+ 5}=5 \) of the circle
Now, equation of the tangent at (a, b) to the given circle is
xa + yb - 2 (x+ a) - 4(y + b)=0
⇒ (a - 2)x + (b - 4)y - (2a + 4b + 5) = 0
If it represents the given line 3x - 4y - k = 0.
Then, \({a-2\over 3}={b-4\over -4}={2a+4b+5\over k}=l\)
⇒ a = 3l + 2, b = 4 - 4l and 2a + 4b + 5 = kl
⇒ 2(3l + 2) + 4(4 - 4l) + 5 = 15l
⇒ l=1 ⇒ a = 5, b = 0
∴ k + a + b = 15 + 5 + 0 = 20
Let \({ X }^{ 2 }+{ Y }^{ 2 }+2gX+2fY+c=0\) be an equation of circle.
Column I | Column II |
A. If circle lies in first quadrant, then | p. g < 0 |
B. If circle lies above X-axis, then | q. g > 0 |
C. If circle lies on the left of Y-axis, then | r. g2 - c < 0 |
D. If circle touches positive X-axis and does not intersect Y.axis then | s. c > 0 |
- (a)
A B C D prs rs qs ps - (b)
A B C D pqr sq pr ps - (c)
A B C D pr ps pq qrs - (d)
None of the above
A. If circle lines in first quadrant, then centre ( -g, -f) lies in first quadrant, thus g < 0 and f < 0; also X and Y-axes must not cut the circle. solving circle and X-axis, we have X2 + 2gX + c = 0, which must have imaginary roots, i.e., g2 - c < 0, then c must be positive. Also, f2 - c < 0.
B. If circle lies above X-axis, then X2 + 2gX + c = 0 must have imaginary roots, then g2 - c < 0 and c > 0.
C. If (-g, -f) lies in second or third quadrant, then g > 0. Also, Y-axis must not cut the circle. Solving circle and Y-axis, we have Y2 + 2fY + c = 0, which must have imaginary roots, i.e., f2 - c < 0, then c must be positive.
D. If circle touches X-axis then X2 + 2gX + c = 0 must have equal roots, i.e., g2 = c, hence c > 0.
Also, \(-g>0\Rightarrow g<0\)
Hence, \(A\rightarrow p,r,s;B\rightarrow r,s;C\rightarrow q,s;D\rightarrow p,s\)
What is the length of an equilateral triangle inscribed in the circle \({ x }^{ 2 }+{ y }^{ 2 }=\frac { 4 }{ 3 } ?\)
- (a)
2 units
- (b)
5 units
- (c)
3 units
- (d)
7 units
Given equation of circle is \({ x }^{ 2 }+{ y }^{ 2 }=\frac { 4 }{ 3 } \)
Radius of circle = \(\frac { 2 }{ \sqrt { 3 } } \)
\(\therefore \) Length of equilateral triangle inscribed in circle
\(=\sqrt { 3 } \times \frac { 2 }{ \sqrt { 3 } } =\) 2 units
\(\because \quad R=\frac { { a }^{ 3 } }{ 4\triangle } \) , where a is the side of equilateral triangle and \(\triangle \) be the area of equilateral triangle.
The cartesian equations of the curves \(X=7+4\cos { \alpha } \) and \(Y=-3+4\sin { \alpha } \) is
- (a)
\({ x }^{ 2 }+{ y }^{ 2 }-14x+6y+42=0\)
- (b)
\({ x }^{ 2 }+{ y }^{ 2 }-6x+14y+21=0\)
- (c)
\({ x }^{ 2 }+{ y }^{ 2 }-10x+12y+28=0\)
- (d)
None of the above
As per the parametric equation of the circle, h = 7, k = - 3, r = 4
\(\Rightarrow { \left( X-7 \right) }^{ 2 }+{ \left( Y+3 \right) }^{ 2 }=16\)
\(\Rightarrow { X }^{ 2 }+{ Y }^{ 2 }-14X+6Y+49+9=16\)
\(\Rightarrow { X }^{ 2 }+{ Y }^{ 2 }-14X+6Y+49+9-16=0\)
\(\Rightarrow { X }^{ 2 }+{ Y }^{ 2 }-14X+6Y+42=0\)
The equation of the circle passing through the points (1,0) and (0,1) and having the smallest radius is
- (a)
\({ x }^{ 2 }+{ y }^{ 2 }+x+y-2=0\)
- (b)
\({ x }^{ 2 }+{ y }^{ 2 }-2x-2y+1=0\)
- (c)
\({ x }^{ 2 }+{ y }^{ 2 }-x-y=0\)
- (d)
\({ x }^{ 2 }+{ y }^{ 2 }+2x+2y-7=0\)
Circle whose diametric end points are (1,0) and (0,1) will be of smallest radius.
\(\Rightarrow \quad \left( X-1 \right) \left( X-0 \right) +\left( Y-0 \right) \left( Y-1 \right) =0 \)
\(\Rightarrow \quad { X }^{ 2 }+{ Y }^{ 2 }-X-Y=0\)
The point diametrically opposite to the point P(1,0) on the circle \({ X }^{ 2 }+{ Y }^{ 2 }+2X+4Y-3=0\) is
- (a)
(3, 4)
- (b)
(3, -4)
- (c)
(-3, 4)
- (d)
(-3, -4)
Given equation can be rewritten as \({ \left( X+1 \right) }^{ 2 }+{ \left( Y+1 \right) }^{ 2 }={ \left( 2\sqrt { 2 } \right) }^{ 2 }\) Let required point be \(Q\left( \alpha ,\beta \right) \).
Then, mid-point of P(1,0) and \(Q\left( \alpha ,\beta \right) \) be the centre of the circle.
i.e., \(\frac { \alpha +1 }{ 2 } =-1\)
and \(\frac { \beta +0 }{ 2 } =-2\)
\(\Rightarrow \quad \alpha =-3\quad and\quad \beta =-4\)
So, the required point is (-3, -4)
If a circle passes through the point (a,b) and cuts the circle \({ X }^{ 2 }+{ Y }^{ 2 }={ p }^{ 2 }\) orthogonally, then the equation of the locus of its centre is
- (a)
\(2ax+2by-\left( { a }^{ 2 }+{ b }^{ 2 }+{ p }^{ 2 } \right) =0\)
- (b)
\({ x }^{ 2 }+{ y }^{ 2 }-2ax-3by+\left( { a }^{ 2 }-{ b }^{ 2 }-{ p }^{ 2 } \right) =0\)
- (c)
\(2ax+2by-\left( { a }^{ 2 }+{ b }^{ 2 }+2{ b }^{ 2 }+{ p }^{ 2 } \right) =0\)
- (d)
\({ x }^{ 2 }+{ y }^{ 2 }-3ax-4by+\left( { a }^{ 2 }+{ b }^{ 2 }-{ p }^{ 2 } \right) =0\)
Let the equation of circle be
x2 + y2 + 2gx + 2fy + c =0 it cuts the circle
x2 + y2 = p2 orthogonally.
∴ 2g(0) + 2f(0) = c - p2 ⇒ c = p2
Also, it passes through (a, b)
∴ a2 + b2 + 2ga + 2fb + p = 0
So, locus of (-g, -f) is
a2 + b2 - 2ax - 2by + p2 = 0
⇒ 2ax + 2by - (a2 + b2 + p2) = 0
If a circle having centre at \(\left( \alpha ,\beta \right) \) radius r completely lies with in two lines x + y = 2 and x + y = - 2 , then min \(\left( \left| \alpha +\beta +2 \right| ,\left| \alpha +\beta -2 \right| \right) \) is
- (a)
greater than \(\sqrt { 2 } r\)
- (b)
less than \(\sqrt { 2 } r\)
- (c)
greater than 2r
- (d)
less than 2r
Minimum distance of the centre from line> radius of circle
i.e., \(\left\{ \frac { \left| \alpha +\beta +2 \right| }{ \sqrt { 2 } } ,\frac { \left| \alpha +\beta -2 \right| }{ \sqrt { 2 } } \right\} >r\)
or min\(\left( \left| \alpha +\beta +2 \right| ,\left| \alpha +\beta -2 \right| \right) \) > \(\sqrt { 2 } r\)
If \({ \lambda x }^{ 2 }+{ \mu y }^{ 2 }+\left( \lambda +\mu -4 \right) xy-\lambda x-\mu y-20=0\) represents a circle, the radius of the circle is
- (a)
\(\sqrt { 21 } /2\)
- (b)
\(\sqrt { 42 } /2\)
- (c)
\(2\sqrt { 21 } \)
- (d)
\(\sqrt { 22 } \)
Since the given equation represents a circle
:. The coefficient of x2 = coefficient of y2
\(\lambda =\mu \)
and the coefficient of x y = 0⇒\(\lambda +\mu -4=0\)⇒\(\lambda =\mu =2\)
Thus the equation of the circle becomes
x2 + y2 - x - y - 10 = 0
whose radius is =\(\sqrt { \left\{ { \left( \frac { 1 }{ 2 } \right) }^{ 2 }+{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }+10 \right\} } \)
=\(\sqrt { 42 } /2\)
The set of values of 'c' so that the equations y=\(\left| x \right| \)+c and x2+y2-8\(\left| x \right| \)-9=0 have no solution is
- (a)
\(\left( -\infty ,-3 \right) \cup \left( 3,\infty \right) \)
- (b)
(-3,3)
- (c)
\(\left( -\infty ,-5\sqrt { 2 } \right) \cup \left( 5\sqrt { 2 } ,\infty \right) \)
- (d)
\(\left( 5\sqrt { 2 } -4,\infty \right) \)
Since y=\(\left| x \right| \) +c and x2+y2-8\(\left| x \right| \)-9=0 both are symmetrical about y-axis for x > 0, y = x + c.
Equation of tangent to circle x2 + y2 - 8x - 9 = 0
Parallel to y = x + c is y = (x - 4) + 5√1+1
for no solution c > 5√2-4 ∴ C∈\(\left( 5\sqrt { 2 } -4,\infty \right) \)
The lines 2x - 3y = 5 and 3x - 4y = 7 are the diameters of a circle of area 154 sq unit. The equation of this circle is
- (a)
x2+y2+2x-2y=62
- (b)
x2+y2+2x-2y=47
- (c)
x2+y2-2x-2y=47
- (d)
x2+y2-2x-2y=62
The centre of the circle is the point of intersection of the given diameters 2x - 3y = 5 and 3x - 4y = 7. Which is (1, -1) and the radius is r, where πr2 =154⇒r2=154 x \(\frac { 7 }{ 22 } \)⇒r=7
and hence the required equation of the circle is
(x - 1)2 + (y + 1)2 = 72
⇒ x2 + y2 - 2x + 2y = 47
The number of rational point(s) (a point (a, b) is rational, if a and b bath are rational numbers) on the circumference of a circle having centre (ㅠ,e) is
- (a)
at most one
- (b)
at least two
- (c)
exactly two
- (d)
infinite
Radius = \(\sqrt { { \left( a-\pi \right) }^{ 2 }{ \left( b-e \right) }^{ 2 } } \)
= irrational = k
Circle \({ \left( x-\pi \right) }^{ 2 }{ \left( y-e \right) }^{ 2 }={ k }^{ 2 }\)
AB is a diameter of a circle and C is any point on the circumference of the circle. Then
- (a)
the area of ∆ABC is maximum when it is isosceles
- (b)
the area of ∆ABC is minimum when it is isosceles
- (c)
the perimeter of ∆ABC is maximum when it is isosceles
- (d)
none of the above
Area of triangle=\(\frac{1}{2}\)(base)(height)
Area of triangle is maximum if height of the triangle is maximum. We know that height is maximum of the isosceles triangle.
Hence area of ∆ABC is maximum when it is isosceles
The four points of intersection of the lines (2x - y + 1) (x - 2y + 3) = 0 with the axes lie on a circle whose centre is at the point
- (a)
(-7/4,5/4)
- (b)
(3/4,5/4)
- (c)
(9/4,5/4)
- (d)
(0,5/4)
Equation of conic is (2x - Y + l)(x - 2y + 3) +λxy=0
for circle coefficient of .xy = 0
-5+λ=0 ஃλ=5
Circle is 2x2+2y2+7x-5y+3=0
\({ x }^{ 2 }+{ y }^{ 2 }+\frac { 7 }{ 2 } x-\frac { 5 }{ 2 } y+\frac { 3 }{ 2 } =0\)
Centre is (-7/4,5/4)
Three sides of a triangle have the equations Lr≡y-m1x-cr=0; r=1,2,3. Then \(\lambda { L }_{ 2 }{ L }_{ 3 }+\mu { L }_{ 3 }{ L }_{ 1 }+v{ L }_{ 1 }{ L }_{ 2 }=0\) , where \(\lambda \neq 0,\mu \neq 0,v\neq 0\) is the equation of circumcircle of triangle, if
- (a)
\(\lambda \left( { m }_{ 2 }+{ m }_{ 3 } \right) +\mu \left( { m }_{ 3 }+{ m }_{ 1 } \right) +v\left( { m }_{ 1 }+{ m }_{ 2 } \right) =0\)
- (b)
\(\lambda \left( { m }_{ 2 }{ m }_{ 3 }-1 \right) +\mu \left( { m }_{ 3 }{ m }_{ 1 }-1 \right) +v\left( { m }_{ 1 }{ m }_{ 2 }-1 \right) =0\)
- (c)
both (a) and (b)
- (d)
none of the above
Given \(\lambda { L }_{ 2 }{ L }_{ 3 }+\mu { L }_{ 3 }{ L }_{ 1 }+v{ L }_{ 1 }{ L }_{ 2 }=0\)
\(\Rightarrow \lambda \left( y-{ m }_{ 2 }x-{ c }_{ 2 } \right) \left( y-{ m }_{ 3 }x-{ c }_{ 3 } \right) +\mu \left( y-{ m }_{ 3 }x-{ c }_{ 3 } \right) \left( y-{ m }_{ 1 }x-{ c }_{ 1 } \right) +v\left( y-{ m }_{ 1 }x-{ c }_{ 1 } \right) \left( y-{ m }_{ 2 }x-{ c }_{ 2 } \right) =0\)
for circle coefficient of x2 = coefficient of y2 and coefficient of xy = 0,then \(\lambda \left( { m }_{ 2 }{ m }_{ 3 }-1 \right) +\mu \left( { m }_{ 3 }{ m }_{ 1 }-1 \right) +v\left( { m }_{ 1 }{ m }_{ 2 }-1 \right) =0\)and \(\lambda \left( { m }_{ 2 }+{ m }_{ 3 } \right) +\mu \left( { m }_{ 3 }+{ m }_{ 1 } \right) +v\left( { m }_{ 1 }+{ m }_{ 2 } \right) =0\)
The abscissaes of two points A and B are the roots of the equation x2+2ax-b2=0 and their ordinate are the roots of the equation x2+2px-q2=0. The radius of the circle with AB as diameter is
- (a)
\(\sqrt { \left( { a }^{ 2 }+{ b }^{ 2 }+{ p }^{ 2 }+{ q }^{ 2 } \right) } \)
- (b)
\(\sqrt { \left( { a }^{ 2 }+{ p }^{ 2 } \right) } \)
- (c)
\(\sqrt { \left( { b }^{ 2 }+{ q }^{ 2 } \right) } \)
- (d)
none of these
Let coordinates of A and B are (∝,β) and (୪,ბ) respectively.
∝+Y=-2a,∝Y=-b2
and β+δ=-2p, βδ=-q2
Equation of circle with AB as diameter
(x-∝)(x-γ)+(y-β)(y-δ)=0
x2+y2-x(∝+⋎)+(y-β)(y-δ)=0
⇒ x2+y2-x(∝+Y)-y(β+δ)+∝Y+βδ=0
Radius=\(\sqrt { \left( { a }^{ 2 }+{ p }^{ 2 }+{ b }^{ 2 }+{ q }^{ 2 } \right) } \)
=\(\sqrt { \left( { a }^{ 2 }+{ b }^{ 2 }+{ p }^{ 2 }+{ q }^{ 2 } \right) } \)
If the two circles x2+y2+2gx+2fy=0 and x2+y2+2g1x+2f1y=0 touch each other, then
- (a)
f1g=fg1
- (b)
ff1=gg1
- (c)
f2+g2=f12+g12
- (d)
none of these
Centre and radius of the circle x2+y2+2gx+2fy=0 are C1(-g,-f) and r1=\(\sqrt { \left( { g }^{ 2 }+{ f }^{ 2 } \right) } \) and centre and radius of the circle x2+y2+2g1x+2f1y=0 are C2(-g1,-f1)
and r2=\(\sqrt { \left( { { g }_{ 1 } }^{ 2 }+{ { f }_{ 1 } }^{ 2 } \right) } \)
Circles touch each other
\(\left| { C }_{ 1 }{ C }_{ 2 } \right| =\left| { r }_{ 1 }\pm { r }_{ 2 } \right| \)
Squaring both sides then
g2 + f2 + g12 + f12 - 2gg1 - 2ff1 = g2 + f2 + g12+ f12 \(\pm\) 2\(\sqrt { \left( { g }^{ 2 }+{ f }^{ 2 } \right) } \)\(\sqrt { \left( { { g }_{ 1 } }^{ 2 }+{ { f }_{ 1 } }^{ 2 } \right) } \)
(gg1+ff1)2=(g2+f2)(g12+f12)
g2g12+f2f12+2gg1ff1=g2g12+g2f12+f2g12+f2f12
g2f12+f2g12-2gg1ff1=0
(gf1-fg1)2=0
gf1=fg1
One of the diameter of the circle x2+y2-12x+4y+6=0 is given by
- (a)
x+y=0
- (b)
x+3y=0
- (c)
x=y
- (d)
3x+2y=0
Diameter always passes through the centre of circle.
If one circle of a coaxial system is x2+y2+2gx+2fy+c=0 and one limiting point is (a, b), then equation of the radical axis will be
- (a)
(g + a)x + (f + b)y + c - a2 - b2 = 0
- (b)
2(g+a)x+2(f+b)y+c-a2-b2=0
- (c)
2gx+2fy+c-a2-b2=0
- (d)
none of the above
Given circle is x2+y2+2gx+2fy+c=0...(i)
end the circle with limiting point (a, b) is
(x-a)2+(y-b)2=0
or x2+y2-2ax-2by+a2+b2=0
or x2+y2-2ax-2by+a2+b2=0...(ii)
Equation of radical axis is
2(g+a)x+2(f+b)y+c-a2-b2=0
S≡x2+y2+2x+3y+1=0 and S'≡x2+y2+4x+3y+2=0 are two circles. The point (-3, - 2) lies
- (a)
inside S' only
- (b)
inside S only
- (c)
inside S and S'
- (d)
outside S and S'
S( - 3, - 2) = 9 + 4 - 6 - 6 + 1
=2>0
(-3, - 2) outside of S
and S' (-3, - 2) = 9 + 4 -12 - 6 + 2 = - 3 < 0
(-3, -2) inside of S'
A circle of radius 5 unit touches both the axes and lies in the first quadrant. If the circle makes one complete roll on x-axis along the positive direction of x-axis, then its equation in the new position is
- (a)
x2 + y2 + 20πx - 10y + 100 π2= a
- (b)
x2 + y2 + 20πx + 10y + 100 π2= a
- (c)
x2 + y2 - 20πx - 10y + 100 π2= a
- (d)
none of the above
The x-coordinate of the centre of the new position of the circle
= 5+ circumference of the first circle
=5+10π
and y coordinate of the centre of the new position of the circle =5
and radius is also 5.
Equation of circle is (x-5-10π)2+(y-5)2=25
If two circles (x-1)2+(y-3)2=r2 and x2+y2-8x+2y+8=0 intersect in two distinct points, then
- (a)
2<r<8
- (b)
r < 2
- (c)
r=2
- (d)
r > 2
Centres and radii of the given circles are C1 (1,3), Ii. = r and C2(4, -1), r2 = 3 respectively since circles intersect in two distinct points, then
\(\left| { r }_{ 1 }-{ r }_{ 2 } \right| <{ C }_{ 1 }{ C }_{ 2 }<{ r }_{ 1 }+{ r }_{ 2 }\)
\(\left| r-3 \right| <5<r+3\)...(i)
From last two relations, r> 2
From first two relations
\(\left| r-3 \right| <5\)
-5<r-3<5
-2 < r <8...(ii)
From Eqs. (i) and (ii), we get 2 < r < 8
The locus of the centres of the circles which cut the circles x2 + y2 + 4x - 6y + 9 = a and x2 + y2 - 5x + 4y + 2 = a orthogonally is
- (a)
3x + 4y - 5 = 0
- (b)
9x - 10y + 7 = 0
- (c)
9x + 10y - 7 = 0
- (d)
9x - 10y + 11 = 0
Let circle x2 + y2 + 2gx + 2fy + c = 0 according to question 2g x 2+2f x -3=c+9
4g - 6f = c + 9...(i)
and 2g x -\(\frac{5}{2}\)+ 2f x 2 = c + 2
-5g+4f = c+2...(ii)
Subtracting Eqs. (ii) from (i),
9g - 10f = 7
-9( -g) + 10(-f) =7
ஃ Locus of centre is -9x + 10y =7 or 9x -10y + 7 = 0
If 4l2 - 5m2 + 6l+ 1 = a and the line lx + my + 1 = 0 touches a fixed circle, then
- (a)
the centre of the circle is at the point (4, 0)
- (b)
the radius of the circle is equal to √5
- (c)
the circle passes through origin
- (d)
none of the above
Let the circle be (x - h)2 + (y - k)2 = r2
then \(r=\frac { \left| lh+mk+1 \right| }{ \sqrt { \left( { l }^{ 2 }+{ m }^{ 2 } \right) } } \)
or l2(h2 - r2) + m2(k2 - r2) + 2lmhk + 2lh + 2mk + 1 = 0
also 4 l2 - 5m2 + 6l + 1 = 0
Comparing the coefficients of similar terms
k = 0, h = 3 and 9 - r2 = 4
r=√5
Hence, centre (3, 0) and radius √5
The circle passing through the distinct points (1, t), (t, 1) and (t, t) for all values of t, passes through the point
- (a)
(1,1)
- (b)
(-1,-1)
- (c)
(1,-1)
- (d)
(-1,1)
Let circle x2 + y2 + 2gx + 2fy + c = 0...(A)
its passes through (1, t ), (t, 1) and (t, t ) , then 1 + t2 + 2g + 2ft + c = 0...(i)
t2 + 1 + 2gt + 2ft + c = 0...(ii)
2t2 + 2gt + 2ft + c = 0...(iii)
Subtracting Eq. (i) from (ii) and Eq. (ii) from (iii)
then 2g(t -1)+ 2f(1-t)=0
or g-f=0
and t2-1+2f(t-1)=0
f=-\(\frac { \left( t+1 \right) }{ 2 } =g\)
From Eq. (iii), 2t2 - t (t + 1) - t (t + 1) + c = 0 c=2t
From (A),
x2 + y2 - (t + 1) x - (r + 1) y + 2t =0
P+λQ=0
P = 0 and Q = 0,
then x2 + y2 - x - y = 0 and x + y - 2 = 0
after solving we get x = 1 and y = 1
Equation of the normal to the circle x2 + y2 - 4x + 4y - 17 = 0 which passes through (1, 1) is
- (a)
3x + 2y - 5 = 0
- (b)
3x + y - 4 = 0
- (c)
3x + 2y - 2 = 0
- (d)
3x - y - 8 = 0
Normal of circle always pass through the centre of the circle
ie., (2, - 2) but its also pass through (1, 1)
Equation of normal
y+2=\(\frac { 1+2 }{ 1-2 } \left( x-2 \right) \)
or - y - 2 = 3x - 6
or 3x + y - 4 = 0
∝, β and ૪ are parametric angles of three points P, Q and R respectively, on the circle x2 + y2 = 1and A is the point (-1, 0). If the lengths of the chords AP, AQ and AR are in GP, then cos ∝/2, cos β/2 and cos ૪/2 are in
- (a)
AP
- (b)
GP
- (c)
HP
- (d)
none of these
Coordinates of P, Q, R are (cos∝, sin∝),(cosβ, sinβ) and (cos૪, sin૪) respectively.
and A≡(-1,0)
AP=\(\sqrt { { \left( 1+cos\alpha \right) }^{ 2 }+{ sin }^{ 2 }\alpha } =2cos\alpha /2\)
AQ=\(\sqrt { { \left( 1+cos\beta \right) }^{ 2 }+{ sin }^{ 2 }\beta } =2cos\beta /2\)
AR=\(\sqrt { { \left( 1+cos\gamma \right) }^{ 2 }+{ sin }^{ 2 }\gamma } =2cos\gamma /2\)
AP, AQ, AR are in GP then
cos ∝/2, cosβ/2, cosY/2 are also in G.P
The equation of the circle passing through (2, 0) and (0,4) and having the minimum radius is
- (a)
x2 + y2 = 20
- (b)
x2 + y2 - 2x - 4y = 0
- (c)
(x2 + y2 - 4) + A(x2 + y2 -16) = 0
- (d)
none of the above
Centre C≡(7,5) and radius r=\(\sqrt { (49+25+151) } \)
=15
If P (2,-7)
The shortest distance = \(\left| CP-r \right| \)
=\(\left| \sqrt { 25+144 } -15 \right| \)
=\(\left| 13-15 \right| =2\)
Equation of the circle cutting orthogonally the three circles x2 + y2 - 2x + 3y - 7 = 0, x2 + y2 + 5x - 5y + 9 = 0 and x2 + y2 + 7 x - 9y + 29 = 0 is
- (a)
x2 + y2 - 16x -18y - 4 = 0
- (b)
x2+y2-7x+11y+6=0
- (c)
x2 + y2 + 2x - 8y + 9 = 0
- (d)
none of the above
Let equation of circle is x2 + y2 + 2gx + 2fy + c = 0
According to question
2g x -1+ 2f x \(\frac{3}{2}\)=c-7
-2g + 3f = c - 7...(i)
2g x \(\frac{5}{2}\)+2f x -\(\frac{5}{2}\)=c+9
5g-5f=c+9...(ii)
and 2g x \(\frac{7}{2}\)+2f x -\(\frac{9}{2}\)=c+29
⇒ 7g - 9f = c + 29...(iii)
Solving Eqs. (i), (ii), and (iii), we get
g = - 8, f = - 9, c = - 4
Equation of circle is
x2 + y2 - 16x -18y - 4 = 0
A line meets the coordinate axes in A and B. A circle is circumscribed about the triangle OAB. If m and n are the distances of the tangent to the circle at the origin from the points A and B respectively, the diameter of the circle is
- (a)
m(m + n)
- (b)
(m + n)
- (c)
n(m + n)
- (d)
\(\frac{1}{2}\)(m+n)
Let line is \(\frac { x }{ a } +\frac { y }{ b } =1\)
A (a, 0) and B (0, b)
Equation of circle since AB is diameters is
(x - a)(x - 0) + (y - 0 )(y - b) = 0
x2+y2-ax-by=0
Tangent at (0, 0) is
ax+by=0
AM=m=\(\frac { { a }^{ 2 } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
and BN=n=\(\frac { { b }^{ 2 } }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } } } \)
⇒ m+n=\(\sqrt { { a }^{ 2 }+{ b }^{ 2 } } \)=diameter of circle
The equation of the circle having its centre on the line x + 2y - 3 = 0 and passing through the points of intersection of the circles,x2+ y2 - 2x - 4y + 1 = a and x2 + y2 - 4x - 2y + 1= 0 is
- (a)
x2 + y2 - 6x + 1 = 0
- (b)
x2 + y2 - 3x + 4 = 0
- (c)
x2 + y2 - 2x - 2y + 1 = 0
- (d)
x2 + y2 + 2x - 4y + 4 = 0
Equation of circle through points of intersection of the circles x2 + y2 - 2 x - 4 y + 1 = 0 and x2 + y2 - 4 x - 2 y + 1 = 0 is
(x2 + y2 - 2 x - 4y + 1)+ ⋋(x2 + y2 - 4 x - 2y + 1)= 0
⇒ (1+⋋)x2+(1+λ)y2-2(1+2⋋)x-2(2+λ)y+(1+⋋)=0
⇒ x2+y2-\(\frac { 2(1+2\lambda ) }{ (1+\lambda ) } x-\frac { 2(2+\lambda ) }{ (1+\lambda ) } y+1=0\)...(i)
Centre☰ \(\left( \frac { 1+2\lambda }{ 1+\lambda } ,\frac { 2+\lambda }{ 1+\lambda } \right) \)
Centre lie on x+2y-3=0
5+4⋋-3(1+⋋)=0
⋋=-2
From Eqs.(i),
x2+y2-6x+1=0
If P is a point on the circle x2 + y2 = 9 Q is a point on the line 7x + y + 3 = 0, and the line x - y + 1 = 0, is the perpendicular bisector of PQ, then the coordinates of Pare
- (a)
(3,0)
- (b)
\(\left( \frac { 72 }{ 25 } ,-\frac { 21 }{ 25 } \right) \)
- (c)
(0,3)
- (d)
\(\left( -\frac { 72 }{ 25 } ,\frac { 21 }{ 25 } \right) \)
Let coordinates of P ≡ (3cos ∝, 3 sin ∝)
Let x -coordinate of Q is x1 then
y -coordinate of Q is - 7x1 - 3
Q≡(x1 ,-7x1-3)
x - y + 1 = 0 is the perpendicular bisector of PQ, then mid point of PQ lie on x - y + 1 = 0
⇒ 8x1+3 cos∝-3 sin∝+5=0
⇒ 24x1+9 cos∝-9sin∝+15=0...(i)
& slope of
(x- y + 1=0)x slope of PQ =-1
\(1\times \frac { 3sin\alpha +{ 7x }_{ 1 }+3 }{ 3cos\alpha -{ x }_{ 1 } } =-1\)
3 sin∝+7x1+3=-3 cos∝+x1
6x1+3sin∝+3 cos∝+3=0
24x1 +12sin∝+12cos∝+12=0...(ii)
subtracting (i) & (ii), we obtain
-3 cos∝-21 sin∝+3=0
(1-cos∝)=7 sin∝
(1-cos∝)2=49(1-cos2∝)
cos∝=1 & cos∝=-\(\frac { 24 }{ 25 } \)
Coordinates of Pare (3,0) and \(\left( -\frac { 72 }{ 25 } ,\frac { 21 }{ 25 } \right) \)
The tangents drawn from the origin to the circle x2 + y2 - 2px - 2qy + q2 = 0 are perpendicular, if
- (a)
p=q
- (b)
p2=q2
- (c)
q=-p
- (d)
p2+q2=1
x2 + y2 - 2px - 2qy + q2 = 0
or (x-p)2+(y-q)2=p2...(i)
since tangents are perpendicular, then locus of point of intersection of tangents is director circle.
Director circle of (i) is
(x - p)2 + (y -q)2 = p2 + p2 ...(ii)
since point of intersection of tangent is (0,0), then from Eq.(ii)
p2+q2=2p2
p2=q2
or p=土q
If two circles (x-1)2+(y-3)2=r2 and x2+y2-8x+2y+8=0 intersect in two distinct points, then
- (a)
2
- (b)
r<2
- (c)
r=2
- (d)
r>2
Centres and radii of the given circles are C1 (1,3), Ii. = r and C2(4, -1), r2 = 3 respectively since circles intersect in two distinct points, then
\(\left| { r }_{ 1 }-{ r }_{ 2 } \right| <{ C }_{ 1 }{ C }_{ 2 }<{ r }_{ 1 }+{ r }_{ 2 }\)
\(\left| r-3 \right| <5...(i)
From last two relations, r> 2
From first two relations
\(\left| r-3 \right| <5\)
-5 -2 < r <8...(ii)
From Eqs. (i) and (ii), we get 2 < r < 8
If point P(x, y) is called a lattice point, if x, y ∈I. Then the total number of lattice points in the interior of the circle x2 + y2 = a2 , a ≠ 0 cannot be
- (a)
202
- (b)
203
- (c)
204
- (d)
205
Given circle is x2 + y2 = a2...(i)
clearly (0, 0) will belong the interior of circle (i) Also other points interior to circle (i) will have the coordinates of the form
(士⋋,0),(0士⋋), where ⋋2<a2
and (土⋋,士μ) and (士μ,土⋋) where⋋2+μ2=a2
and ⋋,μ∈I
Number of lattice points in the interior of the circle will be of the form 1 + 4 r + 8t, where r, t = 0,1, 2, ....
Number of such points must be of the form 4 n + 1, where n = 0,1,2, ....
The circles x2 + y2 - 4x - 81 = 0, x2 + y2 + 24x - 81 = 0 intersect each other at points A and B. A line through point A meet one circle at P and a parallel line through B meet the other circle at Q. Then the locus of the mid point of PQ is
- (a)
(x + 5)2 + (y + 0)2 = 25
- (b)
(x - 5)2 + (y - 0)2 = 25
- (c)
x2+y2+10x=0
- (d)
x2+y2-10x=0
Solving x2+ y2 - 4 x - 81 = 0...(i)
and x2+ y2+ 24 x - 81 = 0...(ii)
we get x = 0
and y=±9
A (0, - 9) and B (0, 9)
Equation of line through A (0, - 9) is
y + 9 = m x ~ y = mx - 9...(iii)
Solving Eqs. (i) and (iii), then
x2 (1 + m2) - (18 m + 4 )x = 0
and equation of line through B (0, 9) and parallal to (iii) is
y=mx+⋋⇒9=0+⋋⇒y=mx+9...(iv)
Solving Eqs. (ii) and (iv), then
x2(1 + m2)+ (18m+ 24) x = 0
Let mid point of PQ is (h, k), then
2h=-20/1+m2...(v)
and 2k=-20m/1+m2...(vi)
From Eqs. (v) and (vi),k/h=m
Substituting the value of m in (v)
h2 + k2 = - 10 h
Locus of mid point of PQ is
(x + 5)2 + (y + 0)2 = 25
The equation of a circle is S1≡x2+y2=1. The orthogonal tangents to S1 meet at another circle S2 and the orthogonal tangents to S2 meet at the third circle S3 Then
- (a)
radius of S2 and S3 are in the ratio 1:√2
- (b)
radius of S2 and S3 are in the ratio 1 : 2
- (c)
the circles S1 , S2 and S3 are concentric
- (d)
none of the above
Orthogonal tangents to a circle meet at the director circle
S2≡x2+ y2 = 2·1
S2≡x2+ y2 =2
Also S3≡x2+ y2 =4
Ratio of radius of S2 and S3=1:√2
Also the three circles are concentric.
If the line 3x - 4y -⋋=0 touches the circle x2+y2-4x - 8y - 5 = 0 at (a, b), then ⋋+a+b is equal to
- (a)
20
- (b)
22
- (c)
-30
- (d)
-28
Since the given line touches the given circle, the length of the perpendicular from the centre (2, 4) of the circle from the line 3x-4y-⋋=0 is equal to the radius \(\sqrt { \left( 4+16+5 \right) } =5\) of the circle.
⇒ \(\frac { \left( 3\times 2-4\times 4-\lambda \right) }{ \sqrt { \left( 9+16 \right) } } =\pm 5\Rightarrow \lambda =15\quad or\quad -35\)
Now, equation of the tangent at (a, b) to the given circle is
xa + yb - 2(x + a) - 4(y + b) - 5 = 0
⇒ (a - 2)x + (b - 4)y - (2 a + 4b + 5) = 0
If it represents the given line 3x - 4y -⋋=0, then \(\frac { a-2 }{ 3 } =\frac { b-4 }{ -4 } =\frac { 2a+4b+5 }{ \lambda } =l\)
Then a = 3l + 2, b = 4 - 4l and 2a + 4b + 5 = ⋋l
Again, if λ=-35
⋋+a+b=-35-1+8=-28
Let A ≡ (a, 0) and B ≡ (-a, 0) be two fixed points ∀ a ∈ (-∞, 0) and P moves on a plane such that PA = nPB (n≠ 0).
If |n|≠ 1, then the locus of a point P is
- (a)
a straight line
- (b)
a circle
- (c)
a parabola
- (d)
an ellipse
PA = n(PB), LetP ≡ (x, y)
(PA)2 = n2 (PB)2
(x - a)2 + y2 = n2 ((x + a)2 + y2)...(i)
(n2 -1)x2 + (n2 -1)y2 + (n2 + 1)2ax+(n2-1)a2=0
x2+y2+\(\left( \frac { { n }^{ 2 }+1 }{ { n }^{ 2 }-1 } \right) \)2ax+a2=0...(ii)
which is a circle
Let A ≡ (a, 0) and B ≡ (-a, 0) be two fixed points ∀ a ∈ (-∞, 0) and P moves on a plane such that PA = nPB (n≠ 0).
If n = 1, then the locus of a point P is
- (a)
a straight line
- (b)
a circle
- (c)
a parabola
- (d)
a hyperbola
In Eq. (i), put n = 1, then
(x - a)2 + y2 = (x + a)2 + y2
4ax =0
or x = 0, which is a straight line
Let A ≡ (a, 0) and B ≡ (-a, 0) be two fixed points ∀ a ∈ (-∞, 0) and P moves on a plane such that PA = nPB (n≠ 0).
If n > 1, then
- (a)
A lies inside the circle the circle and B lies outside the circle
- (b)
A lies outside the circle and B lies inside the circle
- (c)
both A and B lies on the circle
- (d)
both A and B lies inside the circle
Let S=x2+y2+\(\left( \frac { { n }^{ 2 }+1 }{ { n }^{ 2 }-1 } \right) \)2ax+a2
SA=a2+0+\(\left( \frac { { n }^{ 2 }+1 }{ { n }^{ 2 }-1 } \right) \)2a2+a2
=2a2\(\left( 1+\frac { { n }^{ 2 }+1 }{ { n }^{ 2 }-1 } \right) \)
=\(\frac { 4{ a }^{ 2 }{ n }^{ 2 } }{ { n }^{ 2 }-1 } >0\)
and SB=a2+0+\(\left( \frac { { n }^{ 2 }+1 }{ { n }^{ 2 }-1 } \right) \)(-2a)2+a2
=2a2\(\left( 1+\frac { { n }^{ 2 }+1 }{ { n }^{ 2 }-1 } \right) \)
=-\(\frac { 4{ a }^{ 2 }{ n }^{ 2 } }{ { n }^{ 2 }-1 } <0\)
Hence A lies outside the circle and B lies inside the circle.
Let A ≡ (a, 0) and B ≡ (-a, 0) be two fixed points ∀ a ∈ (-∞, 0) and P moves on a plane such that PA = nPB (n≠ 0).
If locus of P is a circle, then the circle
- (a)
passes through A and B
- (b)
never passes through A and B
- (c)
passes through A but does not pass through B
- (d)
passes through B but does not pass through A
SA≠ 0
SB ≠ 0
Hence, circle never passes through A and B.
If 7l2 - 9m2 + 81 + 1 = 0 and we have to find equation of circle having Ix + my + 1 = 0 is a tangent and we can adjust given condition as 16l2 + 8l + 1 = 9 (l2 + m2)
or (4l+1)2=9(l2+m2)⇒ \(\frac { \left| 4l+1 \right| }{ \sqrt { \left( { l }^{ 2 }+{ m }^{ 2 } \right) } } =3\)
Centre of circle = (4, 0) and radius = 3 when any two non parallel lines touching a circle, then centre of circle lies on angle bisector of lines.
If x + 2y = 3 and 2x + y = 3 touches a circle whose centre lies on a line 3x + 4y = 5, then possible centres of circle are
- (a)
\(\left( \frac { 5 }{ 3 } ,0 \right) ,\left( 3,-1 \right) \)
- (b)
\(\left( \frac { 5 }{ 7 } ,\frac { 5 }{ 7 } \right) ,\left( 4,-\frac { 3 }{ 4 } \right) \)
- (c)
\(\left( \frac { 5 }{ 3 } ,0 \right) ,\left( 4,-\frac { 3 }{ 4 } \right) \)
- (d)
\(\left( \frac { 5 }{ 7 } ,\frac { 5 }{ 7 } \right) ,\left( 3,-1 \right) \)
x+2y-3=0 and 2x+y-3=0 are non parallel
Bisector of lines
\(\frac { \left( x+2y-3 \right) }{ \sqrt { 5 } } =\pm \frac { \left( 2x+y-3 \right) }{ \sqrt { 5 } } \)
⇒ (x+2y-3)=(2x+y-3)
⇒ x-y=0 ...(i)
and (x+2y-3)=-(2x+y-3)
x+y=2...(ii)
Centre of circle lies on angle bisector of lines.
Also centre lies on a line 3x+4y=5...(iii)
Solving (i) and (iii), we get \(\left( \frac { 5 }{ 7 } ,\frac { 5 }{ 7 } \right) \)
and solving (ii) and (iii), we get (3,-1)
For each natural number k, let C k denotes the circle with radius k units and centre at the origin. On the circle C k ' a particle moves k units in the counter clockwise direction. After completing its motion on Ck> the particle moves to Ck+l in some well defined manner, where l> 0. The motion of the particle continues in this manner.
If k∈N and l = 1, the particle moves in the radial direction from circle Ck to Ck+1 If particle starts from the point (-1, 0), then
- (a)
it will cross the +ve y-axis at (0,4)
- (b)
it will cross the -ve y-axis at (0, - 4)
- (c)
it will cross the +ve y-axis at (0, 5)
- (d)
it will cross the - ve y-axis at (0, - 5)
To cross positive y-axis ShouId cover \(\frac { 3\pi }{ 2 } \) radian
\(\left( \frac { \frac { 3\pi }{ 2 } radian }{ 1radian } \right) =\frac { 3\pi }{ 2 } =\frac { 3 }{ 2 } \times 3.14\)
=4.75
ie., will be on 5th circle
⇒ point (0, 5)
P is a variable point on the line L = 0.Tangents are drawn to the circle x2 + y2 = 4 from P to touch it at Q and R. The parallelogram PQSR is completed.
If P≡(2,3) then the centre of circumcircle of triangle QRS is
- (a)
\(\left( \frac { 2 }{ 13 } ,\frac { 7 }{ 26 } \right) \)
- (b)
\(\left( \frac { 2 }{ 13 } ,\frac { 3 }{ 26 } \right) \)
- (c)
\(\left( \frac { 3 }{ 13 } ,\frac { 9 }{ 26 } \right) \)
- (d)
\(\left( \frac { 3 }{ 13 } ,\frac { 2 }{ 13 } \right) \)
P≡(2,3)
Equation of QR is 2x + 3y = 4
Let S≡(∝,β)
⇒ \(\frac { \alpha -2 }{ 2 } =\frac { \beta -3 }{ 3 } =\frac { -2(4+9-4) }{ (4+9) } =-\frac { 18 }{ 13 } \)
\(\alpha =\frac { 10 }{ -13 } ,\beta =-\frac { 15 }{ 13 } \)
Now, equation of circumcircle of ∆QRS is
(x2 + y2 - 4) + A (2x + 3y - 4) = 0
Circumcentre is (-λ,-\(\frac{-3\lambda}{2}\))≡\(\left( \frac { 3 }{ 13 } ,\frac { 9 }{ 26 } \right) \)