PYQs Learn Notes Mocks Board Questions MHT-CET Maths Circle 71 MHT-CET Maths previous year questions (PYQs) from 66 papers, 2021 to 2026, with answers and worked solutions.
About this chapter Difficulty: 3 easy · 48 moderate · 20 hard. Most-asked subtopics: Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms (22), Tangents — At a Point, With a Given Slope, From an External Point and Their Loci (21), Concentric Circles and Circles Touching a Line or an Axis (10). Updated 13 October 2026. #1 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3 a 3a 3 a is A x 2 + y 2 = 9 a 2 x^2 + y^2 = 9a^2 x 2 + y 2 = 9 a 2 B x 2 + y 2 = 16 a 2 x^2 + y^2 = 16a^2 x 2 + y 2 = 16 a 2 C x 2 + y 2 = 4 a 2 x^2 + y^2 = 4a^2 x 2 + y 2 = 4 a 2 D x 2 + y 2 = a 2 x^2 + y^2 = a^2 x 2 + y 2 = a 2 #2 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The Cartesian coordinates of a point P with parameter θ = 5 π 6 \theta = \dfrac{5\pi}{6} θ = 6 5 π on the circle x 2 + y 2 − 2 x + 4 y + 1 = 0 x^2 + y^2 - 2x + 4y + 1 = 0 x 2 + y 2 − 2 x + 4 y + 1 = 0 are... A B ( 0 , − 2 + 3 ) (0, -2 + \sqrt{3}) ( 0 , − 2 + 3 ) C D ( − 3 + 1 , − 1 ) (-\sqrt{3} + 1, -1) ( − 3 + 1 , − 1 ) #3 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The equation of a circle passing through the foci of the ellipse x 2 25 + y 2 16 = 1 \dfrac{x^2}{25} + \dfrac{y^2}{16} = 1 25 x 2 + 16 y 2 = 1 and having a center at ( 0 , − 4 ) (0, -4) ( 0 , − 4 ) is : A x 2 + y 2 − 8 y − 9 = 0 x^2 + y^2 - 8y - 9 = 0 x 2 + y 2 − 8 y − 9 = 0 B x 2 + y 2 + 8 y − 9 = 0 x^2 + y^2 + 8y - 9 = 0 x 2 + y 2 + 8 y − 9 = 0 C x 2 + y 2 + 8 y − 41 = 0 x^2 + y^2 + 8y - 41 = 0 x 2 + y 2 + 8 y − 41 = 0 D x 2 + y 2 − 8 y − 41 = 0 x^2 + y^2 - 8y - 41 = 0 x 2 + y 2 − 8 y − 41 = 0 MHT-CET · 2026 · 12 May Shift I · Q61 Easy Add #4 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The standard parametric equations of the circle x 2 + y 2 − 4 x − 6 y − 12 = 0 x^2 + y^2 - 4x - 6y - 12 = 0 x 2 + y 2 − 4 x − 6 y − 12 = 0 are ... A x = 2 + 5 cos θ , y = 3 + 5 sin θ x = 2 + 5\cos\theta, y = 3 + 5\sin\theta x = 2 + 5 cos θ , y = 3 + 5 sin θ B x = − 2 + 5 cos θ , y = − 3 + 5 sin θ x = -2 + 5\cos\theta, y = -3 + 5\sin\theta x = − 2 + 5 cos θ , y = − 3 + 5 sin θ C x = 2 + 12 cos θ , y = 3 + 12 sin θ x = 2 + 12\cos\theta, y = 3 + 12\sin\theta x = 2 + 12 cos θ , y = 3 + 12 sin θ D x = 3 + 5 cos θ , y = 2 + 5 sin θ x = 3 + 5\cos\theta, y = 2 + 5\sin\theta x = 3 + 5 cos θ , y = 2 + 5 sin θ #5 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The equation of the circle which passes through the points ( 2 , 3 ) (2, 3) ( 2 , 3 ) and ( 4 , 5 ) (4, 5) ( 4 , 5 ) and whose centre lies on a straight line 4 x − y − 3 = 0 4x - y - 3 = 0 4 x − y − 3 = 0 , is A ( x − 1 ) 2 + ( y − 6 ) 2 = 10 (x - 1)^2 + (y - 6)^2 = 10 ( x − 1 ) 2 + ( y − 6 ) 2 = 10 B ( x − 3 ) 2 + ( y − 4 ) 2 = 2 (x - 3)^2 + (y - 4)^2 = 2 ( x − 3 ) 2 + ( y − 4 ) 2 = 2 C x 2 + ( y − 7 ) 2 = 20 x^2 + (y - 7)^2 = 20 x 2 + ( y − 7 ) 2 = 20 D ( x − 2 ) 2 + ( y − 5 ) 2 = 4 (x - 2)^2 + (y - 5)^2 = 4 ( x − 2 ) 2 + ( y − 5 ) 2 = 4 Done a few? Here’s your next move
#6 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
A circle passes through the point ( 0 , 1 ) (0, 1) ( 0 , 1 ) and touches the parabola y = x 2 y = x^2 y = x 2 at the point ( 1 , 1 ) (1, 1) ( 1 , 1 ) . The centre of the circle is... A ( − 1 2 , − 5 2 ) \left(-\dfrac{1}{2}, -\dfrac{5}{2}\right) ( − 2 1 , − 2 5 ) B ( 1 2 , − 5 2 ) \left(\dfrac{1}{2}, -\dfrac{5}{2}\right) ( 2 1 , − 2 5 ) C ( 1 2 , 5 4 ) \left(\dfrac{1}{2}, \dfrac{5}{4}\right) ( 2 1 , 4 5 ) D ( − 1 2 , 5 4 ) \left(-\dfrac{1}{2}, \dfrac{5}{4}\right) ( − 2 1 , 4 5 ) #7 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
If a circle passes through the points ( 2 , 3 ) (2, 3) ( 2 , 3 ) and ( 4 , 5 ) (4, 5) ( 4 , 5 ) and its center lies on the straight line y − 4 x + 3 = 0 y - 4x + 3 = 0 y − 4 x + 3 = 0 , then its equation is...... A x 2 + y 2 − 4 x − 10 y + 25 = 0 x^2 + y^2 - 4x - 10y + 25 = 0 x 2 + y 2 − 4 x − 10 y + 25 = 0 B x 2 + y 2 − 4 x − 10 y − 25 = 0 x^2 + y^2 - 4x - 10y - 25 = 0 x 2 + y 2 − 4 x − 10 y − 25 = 0 C x 2 + y 2 − 4 x + 10 y − 25 = 0 x^2 + y^2 - 4x + 10y - 25 = 0 x 2 + y 2 − 4 x + 10 y − 25 = 0 D x 2 + y 2 + 25 = 0 x^2 + y^2 + 25 = 0 x 2 + y 2 + 25 = 0 #8 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
If the equation 3 x 2 + ( 3 − p ) x y + q y 2 − 2 p x = 8 p q 3x^2 + (3 - p)xy + qy^2 - 2px = 8pq 3 x 2 + ( 3 − p ) x y + q y 2 − 2 p x = 8 pq represents a circle, then the area (in sq. units) of this circle is #9 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The equation of a circle which passes through the points ( 2 , 3 ) (2, 3) ( 2 , 3 ) and ( 4 , 5 ) (4, 5) ( 4 , 5 ) and whose center lies on the straight line y − 4 x + 3 = 0 y - 4x + 3 = 0 y − 4 x + 3 = 0 is A x 2 + y 2 − 4 x − 10 y + 25 = 0 x^2 + y^2 - 4x - 10y + 25 = 0 x 2 + y 2 − 4 x − 10 y + 25 = 0 B x 2 + y 2 − 4 x − 10 y − 25 = 0 x^2 + y^2 - 4x - 10y - 25 = 0 x 2 + y 2 − 4 x − 10 y − 25 = 0 C x 2 + y 2 − 4 x + 10 y − 25 = 0 x^2 + y^2 - 4x + 10y - 25 = 0 x 2 + y 2 − 4 x + 10 y − 25 = 0 D x 2 + y 2 + 4 x − 10 y + 25 = 0 x^2 + y^2 + 4x - 10y + 25 = 0 x 2 + y 2 + 4 x − 10 y + 25 = 0 #10 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The centre and radius of the circle ( a + 1 ) x 2 + 3 y 2 − 6 x + 9 y + a + 4 = 0 (a+1)x^2 + 3y^2 - 6x + 9y + a + 4 = 0 ( a + 1 ) x 2 + 3 y 2 − 6 x + 9 y + a + 4 = 0 are respectively ... A ( − 1 , 3 2 ) , 5 2 \left(-1, \dfrac{3}{2}\right), \dfrac{\sqrt{5}}{2} ( − 1 , 2 3 ) , 2 5 B ( − 1 , − 3 2 ) , 5 2 \left(-1, -\dfrac{3}{2}\right), \dfrac{\sqrt{5}}{2} ( − 1 , − 2 3 ) , 2 5 C ( 1 , − 3 2 ) , 5 2 \left(1, -\dfrac{3}{2}\right), \dfrac{\sqrt{5}}{2} ( 1 , − 2 3 ) , 2 5 D ( 1 , 3 2 ) , 5 2 \left(1, \dfrac{3}{2}\right), \dfrac{\sqrt{5}}{2} ( 1 , 2 3 ) , 2 5 MHT-CET · 2025 · 20 April Shift II · Q102 Hard Add #11 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The number of integral values of k for which x 2 + y 2 + k x + ( 1 − k ) y + 5 = 0 x^{2}+y^{2}+ kx+ (1 - k)y+ 5 = 0 x 2 + y 2 + k x + ( 1 − k ) y + 5 = 0 represents a circle whose radius cannot exceeds 5 , are #12 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The equation of the circle passing through the point ( 1 , 1 ) (1,1) ( 1 , 1 ) and having two diameters along the pair of lines x 2 − y 2 − 2 x + 4 y − 3 = 0 x^{2}-y^{2}- 2x+ 4y- 3 = 0 x 2 − y 2 − 2 x + 4 y − 3 = 0 is A ( x + 2 ) 2 + ( y − 2 ) 2 = 4 \ (x + 2)^{2} + (y - 2)^{2} = 4 ( x + 2 ) 2 + ( y − 2 ) 2 = 4 B ( x − 3 ) 2 + ( y − 1 ) 2 = 4 \ (x - 3)^{2} + (y - 1)^{2} = 4 ( x − 3 ) 2 + ( y − 1 ) 2 = 4 C ( x − 1 ) 2 + ( y − 2 ) 2 = 1 \ (x - 1)^{2} + (y - 2)^{2} = 1 ( x − 1 ) 2 + ( y − 2 ) 2 = 1 D ( x + 1 ) 2 + ( y + 2 ) 2 = 1 (x + 1)^{2} + (y + 2)^{2} = 1 ( x + 1 ) 2 + ( y + 2 ) 2 = 1 #13 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
If the sides of a rectangle are given by the equations x = − 2 , x = 6 , y = − 2 , y = 5 x=-2, x=6, y=-2, y=5 x = − 2 , x = 6 , y = − 2 , y = 5 , then the equation of the circle, drawn on the diagonal of this rectangle as its diameter, is A x 2 + y 2 + 4 x + 3 y + 22 = 0 x^2+y^2+4x+3y+22=0 x 2 + y 2 + 4 x + 3 y + 22 = 0 B x 2 + y 2 − 4 x + 3 y − 22 = 0 x^2+y^2-4x+3y-22=0 x 2 + y 2 − 4 x + 3 y − 22 = 0 C x 2 + y 2 − 4 x − 3 y − 22 = 0 x^2+y^2-4x-3y-22=0 x 2 + y 2 − 4 x − 3 y − 22 = 0 D x 2 + y 2 + 4 x − 3 y + 22 = 0 x^2+y^2+4x-3y+22=0 x 2 + y 2 + 4 x − 3 y + 22 = 0 #14 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The parametric equations of the circle x 2 + y 2 − a x − b y = 0 x^2+y^2-ax-by=0 x 2 + y 2 − a x − b y = 0 are A x = a 2 + a 2 + b 2 2 cos θ , y = b 2 + a 2 + b 2 2 sin θ x=\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2}\cos\theta,\ y=\frac{b}{2}+\frac{\sqrt{a^2+b^2}}{2}\sin\theta x = 2 a + 2 a 2 + b 2 cos θ , y = 2 b + 2 a 2 + b 2 sin θ B x = − a 2 + a 2 + b 2 4 sin θ , y = − b 2 + a 2 + b 2 4 cos θ x=-\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{4}\sin\theta,\ y=-\frac{b}{2}+\frac{\sqrt{a^2+b^2}}{4}\cos\theta x = − 2 a + 4 a 2 + b 2 sin θ , y = − 2 b + 4 a 2 + b 2 cos θ C x = a 2 + a 2 + b 2 2 sin θ , y = b 2 + a 2 + b 2 2 cos θ x=\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2}\sin\theta,\ y=\frac{b}{2}+\frac{\sqrt{a^2+b^2}}{2}\cos\theta x = 2 a + 2 a 2 + b 2 sin θ , y = 2 b + 2 a 2 + b 2 cos θ D x = a 2 + a 2 + b 2 4 cos θ , y = b 2 + a 2 + b 2 4 sin θ x=\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{4}\cos\theta,\ y=\frac{b}{2}+\frac{\sqrt{a^2+b^2}}{4}\sin\theta x = 2 a + 4 a 2 + b 2 cos θ , y = 2 b + 4 a 2 + b 2 sin θ #15 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The abscissae of the two points A and B are the roots of the equation x 2 + 2 a x − b 2 = 0 x^2+2ax-b^2=0 x 2 + 2 a x − b 2 = 0 and their ordinates are roots of the equation y 2 + 2 p y − q 2 = 0 y^2+2py-q^2=0 y 2 + 2 p y − q 2 = 0 . Then the equation of the circle with AB as diameter is given by A x 2 + y 2 − 2 a x − 2 p y + b 2 + q 2 = 0 x^2+y^2-2ax-2py+b^2+q^2=0 x 2 + y 2 − 2 a x − 2 p y + b 2 + q 2 = 0 B x 2 + y 2 − 2 a x − 2 p y − b 2 + q 2 = 0 x^2+y^2-2ax-2py-b^2+q^2=0 x 2 + y 2 − 2 a x − 2 p y − b 2 + q 2 = 0 C x 2 + y 2 + 2 a x + 2 p y + b 2 + q 2 = 0 x^2+y^2+2ax+2py+b^2+q^2=0 x 2 + y 2 + 2 a x + 2 p y + b 2 + q 2 = 0 D x 2 + y 2 + 2 a x + 2 p y − b 2 − q 2 = 0 x^2+y^2+2ax+2py-b^2-q^2=0 x 2 + y 2 + 2 a x + 2 p y − b 2 − q 2 = 0 #16 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The equation of the circle, the end points of whose diameter are the centres of the circles x 2 + y 2 + 6 x − 14 y + 5 = 0 x^2+y^2+6x-14y+5=0 x 2 + y 2 + 6 x − 14 y + 5 = 0 and x 2 + y 2 − 4 x + 10 y − 4 = 0 x^2+y^2-4x+10y-4=0 x 2 + y 2 − 4 x + 10 y − 4 = 0 is A x 2 + y 2 − x − 2 y + 41 = 0 x^2+y^2-x-2y+41=0 x 2 + y 2 − x − 2 y + 41 = 0 B x 2 + y 2 + x − 2 y − 41 = 0 x^2+y^2+x-2y-41=0 x 2 + y 2 + x − 2 y − 41 = 0 C x 2 + y 2 + x − 2 y − 41 = 0 x^2+y^2+x-2y-41=0 x 2 + y 2 + x − 2 y − 41 = 0 D x 2 + y 2 − x + 2 y − 41 = 0 x^2+y^2-x+2y-41=0 x 2 + y 2 − x + 2 y − 41 = 0 #17 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
If ( m i , 1 m i ) \left(m_i, \dfrac{1}{m_i}\right) ( m i , m i 1 ) , m i > 0 m_i > 0 m i > 0 , i = 1 , 2 , 3 , 4 i = 1, 2, 3, 4 i = 1 , 2 , 3 , 4 are four distinct points on a circle, then the product m 1 m 2 m 3 m 4 m_1 m_2 m_3 m_4 m 1 m 2 m 3 m 4 is equal to #18 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The abscissae of two points A and B are the roots of the equation x 2 + 2 a x − b 2 = 0 x^2+2ax-b^2=0 x 2 + 2 a x − b 2 = 0 and their ordinates are roots of the equation y 2 + 2 p y − q 2 = 0 y^2+2py-q^2=0 y 2 + 2 p y − q 2 = 0 . Then the equation of the circle with AB as diameter is given by A x 2 + y 2 − 2 a x − 2 p y + b 2 + q 2 = 0 x^2+y^2-2ax-2py+b^2+q^2=0 x 2 + y 2 − 2 a x − 2 p y + b 2 + q 2 = 0 B x 2 + y 2 − 2 a x − 2 p y − b 2 − q 2 = 0 x^2+y^2-2ax-2py-b^2-q^2=0 x 2 + y 2 − 2 a x − 2 p y − b 2 − q 2 = 0 C x 2 + y 2 + 2 a x + 2 p y + b 2 + q 2 = 0 x^2+y^2+2ax+2py+b^2+q^2=0 x 2 + y 2 + 2 a x + 2 p y + b 2 + q 2 = 0 D x 2 + y 2 + 2 a x + 2 p y − b 2 − q 2 = 0 x^2+y^2+2ax+2py-b^2-q^2=0 x 2 + y 2 + 2 a x + 2 p y − b 2 − q 2 = 0 #19 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
If a circle passes through points ( 4 , 0 ) (4,0) ( 4 , 0 ) and ( 0 , 2 ) (0,2) ( 0 , 2 ) and its centre lies on Y-axis. If the radius of the circle is r, then the value of r 2 − r + 1 r^2-r+1 r 2 − r + 1 is #20 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
The parametric equation of the circle x 2 + y 2 + 2 x − 4 y − 4 = 0 x^2+y^2+2x-4y-4=0 x 2 + y 2 + 2 x − 4 y − 4 = 0 are A x = − 1 + 3 cos θ , y = 2 + 3 sin θ x=-1+3\cos\theta,\;y=2+3\sin\theta x = − 1 + 3 cos θ , y = 2 + 3 sin θ B x = 1 + 3 cos θ , y = − 2 + 3 sin θ x=1+3\cos\theta,\;y=-2+3\sin\theta x = 1 + 3 cos θ , y = − 2 + 3 sin θ C x = − 1 + 3 sin θ , y = − 2 + 3 cos θ x=-1+3\sin\theta,\;y=-2+3\cos\theta x = − 1 + 3 sin θ , y = − 2 + 3 cos θ D x = 1 + 3 sin θ , y = − 2 + 3 cos θ x=1+3\sin\theta,\;y=-2+3\cos\theta x = 1 + 3 sin θ , y = − 2 + 3 cos θ #21 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
If the sides of a rectangle are given by the equations x = − 2 , x = 6 , y = − 2 , y = 5 x = -2,\ x = 6,\ y = -2,\ y = 5 x = − 2 , x = 6 , y = − 2 , y = 5 , then the equation of the circle, drawn on the diagonal of this rectangle as its diameter, is A x 2 + y 2 + 4 x + 3 y + 22 = 0 x^2 + y^2 + 4x + 3y + 22 = 0 x 2 + y 2 + 4 x + 3 y + 22 = 0 B x 2 + y 2 − 4 x + 3 y − 22 = 0 x^2 + y^2 - 4x + 3y - 22 = 0 x 2 + y 2 − 4 x + 3 y − 22 = 0 C x 2 + y 2 − 4 x − 3 y − 22 = 0 x^2 + y^2 - 4x - 3y - 22 = 0 x 2 + y 2 − 4 x − 3 y − 22 = 0 D x 2 + y 2 + 4 x − 3 y + 22 = 0 x^2 + y^2 + 4x - 3y + 22 = 0 x 2 + y 2 + 4 x − 3 y + 22 = 0 MHT-CET · 2022 · Shift 1 · Q145 Moderate Add #22 · Equation of a Circle — Centre-Radius, General, Diameter and Parametric Forms
If the lines 3 x − 4 y − 7 = 0 3x-4y-7=0 3 x − 4 y − 7 = 0 and 2 x − 3 y − 5 = 0 2x-3y-5=0 2 x − 3 y − 5 = 0 pass through diameters of a circle of area 49 π 49\pi 49 π square units, then the equation of the circle is A x 2 + y 2 − 2 x + 2 y − 47 = 0 x^2+y^2-2x+2y-47=0 x 2 + y 2 − 2 x + 2 y − 47 = 0 B x 2 + y 2 − 2 x + 2 y + 51 = 0 x^2+y^2-2x+2y+51=0 x 2 + y 2 − 2 x + 2 y + 51 = 0 C x 2 + y 2 + 2 x − 2 y − 51 = 0 x^2+y^2+2x-2y-51=0 x 2 + y 2 + 2 x − 2 y − 51 = 0 D x 2 + y 2 + 2 x + 2 y + 47 = 0 x^2+y^2+2x+2y+47=0 x 2 + y 2 + 2 x + 2 y + 47 = 0 #23 · Concentric Circles and Circles Touching a Line or an Axis
The equation of a circle whose center lies on x + 2 y = 0 x + 2y = 0 x + 2 y = 0 and touching the lines 3 x − 4 y + 8 = 0 3x - 4y + 8 = 0 3 x − 4 y + 8 = 0 and 3 x − 4 y − 28 = 0 3x - 4y - 28 = 0 3 x − 4 y − 28 = 0 is A ( x − 2 ) 2 + ( y + 1 ) 2 = 324 (x - 2)^2 + (y + 1)^2 = 324 ( x − 2 ) 2 + ( y + 1 ) 2 = 324 B ( x − 2 ) 2 + ( y − 1 ) 2 = 324 (x - 2)^2 + (y - 1)^2 = 324 ( x − 2 ) 2 + ( y − 1 ) 2 = 324 C 5 ( x − 2 ) 2 + 5 ( y + 1 ) 2 = 324 5(x - 2)^2 + 5(y + 1)^2 = 324 5 ( x − 2 ) 2 + 5 ( y + 1 ) 2 = 324 D 25 ( x − 2 ) 2 + 25 ( y + 1 ) 2 = 324 25(x - 2)^2 + 25(y + 1)^2 = 324 25 ( x − 2 ) 2 + 25 ( y + 1 ) 2 = 324 #24 · Concentric Circles and Circles Touching a Line or an Axis
The number of circles passing through the origin and touching the lines x + y = 1 x + y = 1 x + y = 1 and x − y = 1 x - y = 1 x − y = 1 is … \ldots … #25 · Concentric Circles and Circles Touching a Line or an Axis
The equation of the circle concentric with circle x 2 + y 2 − 6 x + 7 = 0 x^2 + y^2 - 6x + 7 = 0 x 2 + y 2 − 6 x + 7 = 0 and which touches the line x + y + 3 = 0 x + y + 3 = 0 x + y + 3 = 0 is .... A x 2 + y 2 − 6 x + 9 = 0 x^2 + y^2 - 6x + 9 = 0 x 2 + y 2 − 6 x + 9 = 0 B x 2 + y 2 − 6 x − 9 = 0 x^2 + y^2 - 6x - 9 = 0 x 2 + y 2 − 6 x − 9 = 0 C x 2 + y 2 − 6 x + 3 = 0 x^2 + y^2 - 6x + 3 = 0 x 2 + y 2 − 6 x + 3 = 0 D x 2 + y 2 − 6 x − 3 = 0 x^2 + y^2 - 6x - 3 = 0 x 2 + y 2 − 6 x − 3 = 0 #26 · Concentric Circles and Circles Touching a Line or an Axis
If a circle with center ( − 1 , 1 ) (-1, 1) ( − 1 , 1 ) touches the line x + 2 y + 4 = 0 x + 2y + 4 = 0 x + 2 y + 4 = 0 , then the co-ordinates of the point of contact are #27 · Concentric Circles and Circles Touching a Line or an Axis
If a circle with centre at ( − 1 , 1 ) ( - 1,1) ( − 1 , 1 ) touches the line x + 2 y + 4 = 0 x+ 2y+ 4 = 0 x + 2 y + 4 = 0 then the co-ordinates of the point of contact are A ( − 2 , − 1 ) ( - 2, - 1) ( − 2 , − 1 ) B C D #28 · Concentric Circles and Circles Touching a Line or an Axis
The equation of the circle, concentric with the circle 2 x 2 + 2 y 2 − 6 x + 8 y + 1 = 0 2x^2+2y^2-6x+8y+1=0 2 x 2 + 2 y 2 − 6 x + 8 y + 1 = 0 and double of its area is A 2 x 2 + 2 y 2 − 6 x + 8 y + 11 = 0 2x^2+2y^2-6x+8y+11=0 2 x 2 + 2 y 2 − 6 x + 8 y + 11 = 0 B 2 x 2 + 2 y 2 − 6 x + 8 y − 11 = 0 2x^2+2y^2-6x+8y-11=0 2 x 2 + 2 y 2 − 6 x + 8 y − 11 = 0 C 4 x 2 + 4 y 2 − 12 x + 16 y − 21 = 0 4x^2+4y^2-12x+16y-21=0 4 x 2 + 4 y 2 − 12 x + 16 y − 21 = 0 D 4 x 2 + 4 y 2 − 12 x + 16 y + 21 = 0 4x^2+4y^2-12x+16y+21=0 4 x 2 + 4 y 2 − 12 x + 16 y + 21 = 0 MHT-CET · 2023 · 9th May Shift 2 · Q133 Moderate Add #29 · Concentric Circles and Circles Touching a Line or an Axis
The equation of the circle, concentric with the circle x 2 + y 2 − 6 x − 4 y − 12 = 0 x^2+y^2-6x-4y-12=0 x 2 + y 2 − 6 x − 4 y − 12 = 0 and touching the X-axis is A x 2 + y 2 − 6 x − 4 y + 5 = 0 x^2+y^2-6x-4y+5=0 x 2 + y 2 − 6 x − 4 y + 5 = 0 B x 2 + y 2 − 6 x − 4 y + 17 = 0 x^2+y^2-6x-4y+17=0 x 2 + y 2 − 6 x − 4 y + 17 = 0 C x 2 + y 2 − 6 x − 4 y + 9 = 0 x^2+y^2-6x-4y+9=0 x 2 + y 2 − 6 x − 4 y + 9 = 0 D x 2 + y 2 − 6 x − 4 y + 4 = 0 x^2+y^2-6x-4y+4=0 x 2 + y 2 − 6 x − 4 y + 4 = 0 #30 · Concentric Circles and Circles Touching a Line or an Axis
The equation of the circle which passes through the centre of the circle x 2 + y 2 + 8 x + 10 y − 7 = 0 x^2+y^2+8x+10y-7=0 x 2 + y 2 + 8 x + 10 y − 7 = 0 and concentric with the circle 2 x 2 + 2 y 2 − 8 x − 12 y − 9 = 0 2x^2+2y^2-8x-12y-9=0 2 x 2 + 2 y 2 − 8 x − 12 y − 9 = 0 is A x 2 + y 2 − 4 x − 6 y + 77 = 0 x^2+y^2-4x-6y+77=0 x 2 + y 2 − 4 x − 6 y + 77 = 0 B x 2 + y 2 − 4 x − 6 y − 89 = 0 x^2+y^2-4x-6y-89=0 x 2 + y 2 − 4 x − 6 y − 89 = 0 C x 2 + y 2 − 4 x − 6 y + 97 = 0 x^2+y^2-4x-6y+97=0 x 2 + y 2 − 4 x − 6 y + 97 = 0 D x 2 + y 2 − 4 x − 6 y − 87 = 0 x^2+y^2-4x-6y-87=0 x 2 + y 2 − 4 x − 6 y − 87 = 0 #31 · Concentric Circles and Circles Touching a Line or an Axis
The equation of the circle which has its centre at the point ( 3 , 4 ) (3,4) ( 3 , 4 ) and touches the line 5 x + 12 y − 11 = 0 5x+12y-11=0 5 x + 12 y − 11 = 0 is A x 2 + y 2 − 6 x − 8 y + 9 = 0 x^2+y^2-6x-8y+9=0 x 2 + y 2 − 6 x − 8 y + 9 = 0 B x 2 + y 2 − 6 x − 8 y + 25 = 0 x^2+y^2-6x-8y+25=0 x 2 + y 2 − 6 x − 8 y + 25 = 0 C x 2 + y 2 − 6 x − 8 y − 9 = 0 x^2+y^2-6x-8y-9=0 x 2 + y 2 − 6 x − 8 y − 9 = 0 D x 2 + y 2 − 6 x − 8 y − 25 = 0 x^2+y^2-6x-8y-25=0 x 2 + y 2 − 6 x − 8 y − 25 = 0 #32 · Concentric Circles and Circles Touching a Line or an Axis
The equation of the concentric circle, with circle C 1 C_1 C 1 having equation x 2 + y 2 − 6 x − 4 y − 12 = 0 x^2+y^2-6x-4y-12=0 x 2 + y 2 − 6 x − 4 y − 12 = 0 and having double the area of C 1 C_1 C 1 , is A x 2 + y 2 − 6 x − 4 y = 27 x^2+y^2-6x-4y=27 x 2 + y 2 − 6 x − 4 y = 27 B x 2 + y 2 − 6 x − 4 y = 13 x^2+y^2-6x-4y=13 x 2 + y 2 − 6 x − 4 y = 13 C x 2 + y 2 − 6 x − 4 y = 50 x^2+y^2-6x-4y=50 x 2 + y 2 − 6 x − 4 y = 50 D x 2 + y 2 − 6 x − 4 y = 37 x^2+y^2-6x-4y=37 x 2 + y 2 − 6 x − 4 y = 37 #33 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
The equation of the director circle of the circle x 2 + y 2 + 3 x − 8 y − 3 = 0 x^2 + y^2 + 3x - 8y - 3 = 0 x 2 + y 2 + 3 x − 8 y − 3 = 0 is: A 4 x 2 + 4 y 2 + 12 x − 32 y − 12 = 0 4x^2 + 4y^2 + 12x - 32y - 12 = 0 4 x 2 + 4 y 2 + 12 x − 32 y − 12 = 0 B 4 x 2 + 4 y 2 + 12 x − 32 y − 97 = 0 4x^2 + 4y^2 + 12x - 32y - 97 = 0 4 x 2 + 4 y 2 + 12 x − 32 y − 97 = 0 C x 2 + y 2 + 12 x − 32 y − 97 4 = 0 x^2 + y^2 + 12x - 32y - \dfrac{97}{4} = 0 x 2 + y 2 + 12 x − 32 y − 4 97 = 0 D x 2 + y 2 + 3 x − 8 y − 12 = 0 x^2 + y^2 + 3x - 8y - 12 = 0 x 2 + y 2 + 3 x − 8 y − 12 = 0 #34 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
The angle between the tangents drawn from the point ( 1 , 1 ) (1, 1) ( 1 , 1 ) to the circle 2 x 2 + 2 y 2 − 8 x + 4 y + 5 = 0 2x^2 + 2y^2 - 8x + 4y + 5 = 0 2 x 2 + 2 y 2 − 8 x + 4 y + 5 = 0 is … \ldots … #35 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
If the tangent to the circle x 2 + y 2 = 5 x^2 + y^2 = 5 x 2 + y 2 = 5 at point ( 1 , − 2 ) (1, -2) ( 1 , − 2 ) also touches the circle x 2 + y 2 − 8 x + 6 y + k = 0 x^2 + y^2 - 8x + 6y + k = 0 x 2 + y 2 − 8 x + 6 y + k = 0 , then k 5 = ⋅ s ⋅ s ⋅ s \dfrac{k}{5} = \cdot s\cdot s\cdot s 5 k = ⋅ s ⋅ s ⋅ s #36 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
Let PA and PB be the tangent segments drawn from point P ( 6 , 8 ) P(6, 8) P ( 6 , 8 ) to the circle with the centre at origin O. The radius of circle for which the area of quadrilateral PAOB is maximum, is... A B C 5 2 \dfrac{5}{\sqrt{2}} 2 5 D #37 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
A line making equal intercepts on coordinate axes and is tangent to the circle x 2 + y 2 = 4 x^2 + y^2 = 4 x 2 + y 2 = 4 . The length of each intercept made by line on the coordinate axes is ... MHT-CET · 2026 · 18 April Shift I · Q56 Moderate Add #38 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
The area of the region (in sq. unit) bounded by x-axis, the tangent and normal to the circle x 2 + y 2 = 4 x^2 + y^2 = 4 x 2 + y 2 = 4 , drawn at a point ( 1 , 3 ) (1, \sqrt{3}) ( 1 , 3 ) is #39 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
The equations of the tangent to the curve x 2 + y 2 = 10 x^2 + y^2 = 10 x 2 + y 2 = 10 , where the tangent is parallel to the line 2 x + y − 1 = 0 2x + y - 1 = 0 2 x + y − 1 = 0 , are A 2 x + y = ± 2 2x + y = \pm\sqrt{2} 2 x + y = ± 2 B 2 x + y = ± 5 2 2x + y = \pm 5\sqrt{2} 2 x + y = ± 5 2 C 2 x + y = ± 2 2 2x + y = \pm 2\sqrt{2} 2 x + y = ± 2 2 D 2 x + y = ± 3 2 2x + y = \pm 3\sqrt{2} 2 x + y = ± 3 2 #40 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
The angle between the tangents drawn from the origin to the circle ( x − 7 ) 2 + ( y + 1 ) 2 = 25 (x - 7)^2 + (y + 1)^2 = 25 ( x − 7 ) 2 + ( y + 1 ) 2 = 25 is #41 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
Two tangents to the circle x 2 + y 2 = 4 x^{2}+y^{2}= 4 x 2 + y 2 = 4 at the points A and B meet at P ( − 4 , 0 ) P( - 4,0) P ( − 4 , 0 ) . Then the area of quadrilateral PAOB, where ' O ' is the origin is A B 4 3 \frac{4}{\sqrt{3}} 3 4 sq. units C D MHT-CET · 2025 · 25 April Shift I · Q143 Hard Add #42 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
A pair of tangents are drawn to the circle x 2 + y 2 + 6 x − 4 y − 12 = 0 x^{2}+y^{2}+ 6x- 4y- 12 = 0 x 2 + y 2 + 6 x − 4 y − 12 = 0 from a point P ( − 4 , − 5 ) P( - 4, - 5) P ( − 4 , − 5 ) , then the area enclosed between these tangents and the area of the circle is A 25 ( 4 + π 4 ) 25\left( \frac{4 + \pi}{4} \right) 25 ( 4 4 + π ) sq. units B 25 ( 4 + π 2 ) 25\left( \frac{4 + \pi}{2} \right) 25 ( 2 4 + π ) sq. units C 25 ( 4 − π 2 ) \ 25\left( \frac{4 - \pi}{2} \right) 25 ( 2 4 − π ) sq. units D 25 ( 4 − π 4 ) 25\left( \frac{4 - \pi}{4} \right) 25 ( 4 4 − π ) sq. units #43 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
The equations of the tangents to the circle x 2 + y 2 = 36 x^{2}+y^{2}= 36 x 2 + y 2 = 36 which are perpendicular to the line 5 x + y = 2 5x + y = 2 5 x + y = 2 , are A x + 5 y ± 6 26 = 0 x + 5y \pm 6\sqrt{26} = 0 x + 5 y ± 6 26 = 0 B x − 5 y ± 6 26 = 0 x - 5y \pm 6\sqrt{26} = 0 x − 5 y ± 6 26 = 0 C 5 x − y ± 6 26 = 0 5x - y \pm 6\sqrt{26} = 0 5 x − y ± 6 26 = 0 D 5 x + y ± 6 26 = 0 5x + y \pm 6\sqrt{26} = 0 5 x + y ± 6 26 = 0 #44 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
If the tangent and the normal at the point ( 3 , 1 ) (\sqrt{3},1) ( 3 , 1 ) to the circle x 2 + y 2 = 4 x^{2}+y^{2}= 4 x 2 + y 2 = 4 , and the X X X -axis form a triangle, then the area (in sq.units) of this triangle is #45 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
The locus of point of intersection of the tangents to the circle x 2 + y 2 = 16 x^{2}+y^{2}= 16 x 2 + y 2 = 16 , such that the angle between them is 60 ∘ 60^{\circ} 6 0 ∘ , is A x 2 + y 2 = 4 x^{2} + y^{2} = 4 x 2 + y 2 = 4 B x 2 + y 2 = 64 x^{2} + y^{2} = 64 x 2 + y 2 = 64 C x 2 + y 2 = 32 x^{2} + y^{2} = 32 x 2 + y 2 = 32 D x 2 + y 2 = 48 x^{2} + y^{2} = 48 x 2 + y 2 = 48 #46 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
If the tangent at ( 1 , 7 ) (1,7) ( 1 , 7 ) to the curve x 2 = y − 6 x^{2}=y- 6 x 2 = y − 6 touches the circle x 2 + y 2 + 16 x + 12 y + C = 0 x^{2}+y^{2}+ 16x+ 12y+ C = 0 x 2 + y 2 + 16 x + 12 y + C = 0 , then C = C = C = #47 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
One end of the diameter of the circle x 2 + y 2 − 6 x − 5 y − 1 = 0 x^2+y^2-6x-5y-1=0 x 2 + y 2 − 6 x − 5 y − 1 = 0 is ( − 1 , 3 ) (-1,3) ( − 1 , 3 ) , then the equation of the tangent at the other end of the diameter is A 8 x + y − 58 = 0 8x+y-58=0 8 x + y − 58 = 0 B 8 x − 2 y − 52 = 0 8x-2y-52=0 8 x − 2 y − 52 = 0 C 8 x − y − 54 = 0 8x-y-54=0 8 x − y − 54 = 0 D 8 x + 2 y − 60 = 0 8x+2y-60=0 8 x + 2 y − 60 = 0 #48 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
The equation of the tangent to the circle, given by x = 5 cos θ x = 5\cos\theta x = 5 cos θ , y = 5 sin θ y = 5\sin\theta y = 5 sin θ at the point θ = π / 3 \theta = \pi/3 θ = π /3 on it, is A x − 3 y = − 5 x - \sqrt{3}y = -5 x − 3 y = − 5 B x + 3 y = 10 x + \sqrt{3}y = 10 x + 3 y = 10 C 3 x + y = 5 3 \sqrt{3}x + y = 5\sqrt{3} 3 x + y = 5 3 D 3 x − y = 0 \sqrt{3}x - y = 0 3 x − y = 0 MHT-CET · 2023 · 10th May Shift 2 · Q122 Hard Add #49 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then 2 r 2r 2 r equals A P Q ⋅ R S \sqrt{PQ\cdot RS} P Q ⋅ R S B P Q + R S 2 \frac{PQ+RS}{2} 2 P Q + R S C 2 ⋅ P Q ⋅ R S P Q + R S \frac{2\cdot PQ\cdot RS}{PQ+RS} P Q + R S 2 ⋅ P Q ⋅ R S D P Q 2 + R S 2 2 \sqrt{\frac{PQ^2+RS^2}{2}} 2 P Q 2 + R S 2 MHT-CET · 2023 · 3rd May Shift 2 · Q127 Hard Add #50 · Tangents — At a Point, With a Given Slope, From an External Point and Their Loci
The tangent to the circle x 2 + y 2 = 5 x^{2}+y^{2}=5 x 2 + y 2 = 5 at ( 1 , − 2 ) (1,-2) ( 1 , − 2 ) also touches the circle x 2 + y 2 − 8 x + 6 y + 20 = 0 x^{2}+y^{2}-8x+6y+20=0 x 2 + y 2 − 8 x + 6 y + 20 = 0 then the co-ordinates of the corresponding point of contact is Showing the first 50 of 71.
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