Prove that the equation of the circumcircle of the triangle formed by the lines u 1 ≡ a 1 x + b 1 y + c 1 = 0 u 2 ≡ a 2 x + b 2 y + c 2 = 0 u 3 ≡ a 3 x + b 3 y + c 3 = 0 is 1 u 1 1 u 2 1 u 3 a 2 a 3 - b 2 b 3 a 3 a 1 - b 3 b 1 a 1 a 2 - b 1 b 2 a 2 b 3 + a 3 b 2 a 3 b 1 + a 1 b 3 a 1 b 2 + a 2 b 1 = 0 or a 1 2 + b 1 2 u 1 a 2 2 + b 2 2 u 2 a 3 2 + b 3 2 u 3 a 1 a 2 a 3 b 1 b 2 b 3 = 0

We have, u 1 ≡ a 1 x + b 1 y + c 1 = 0 u 2 ≡ a 2 x + b 2 y + c 2 = 0 u 3 ≡ a 3 x + b 3 y + c 3 = 0 Solving u 1 & u 2 , we get a 1 b 2 x + b 1 b 2 y + c 1 b 2 = 0 a 2 b 1 x + b 1 b 2 y + c 2 b 1 = 0 On subtracting, we get x = c 2 b 1 - c 1 b 2 a 1 b 2 - a 2 b 1 , y = 1 b 1 - c 1 - a 1 c 2 b 1 - c 1 b 2 a 1 b 2 - a 2 b 1 ⇒ y = c 1 a 2 - a 1 c 2 a 1 b 2 - a 2 b 1 So, one vertex of triangle is x 1 , y 1 ≡ c 2 b 1 - c 1 b 2 a 1 b 2 - a 2 b 1 , c 1 a 2 - a 1 c 2 a 1 b 2 - a 2 b 1 Similarly, other vertices are x 2 , y 2 ≡ c 3 b 2 - c 2 b 3 a 2 b 3 - a 3 b 2 , c 2 a 3 - a 2 c 3 a 2 b 3 - a 3 b 2 x 3 , y 3 ≡ c 1 b 3 - c 3 b 1 a 3 b 1 - a 1 b 3 , a 3 c 1 - c 3 a 1 a 3 b 1 - a 1 b 3 Equation of circumcircle is x 2 + y 2 x y 1 x 1 2 + y 1 2 x 1 y 1 1 x 2 2 + y 2 2 x 2 y 2 1 x 3 2 + y 3 2 x 3 y 3 1 = 0 ⇒ x 2 + y 2 x y 1 x 1 2 + y 1 2 - x 2 + y 2 x 1 - x y 1 - y 0 x 2 2 + y 2 2 - x 2 + y 2 x 2 - x y 2 - y 0 x 3 2 + y 3 2 - x 2 + y 2 x 3 - x y 3 - y 0 = 0 ⇒ x 1 2 + y 1 2 - x 2 + y 2 x 1 - x y 1 - y x 2 2 + y 2 2 - x 2 + y 2 x 2 - x y 2 - y x 3 2 + y 3 2 - x 2 + y 2 x 3 - x y 3 - y = 0 On putting value of x i , y i , i = 1 , 2 , 3 and simplifying we get 1 u 1 1 u 2 1 u 3 a 2 a 3 - b 2 b 3 a 3 a 1 - b 3 b 1 a 1 a 2 - b 1 b 2 a 2 b 3 + a 3 b 2 a 3 b 1 + a 1 b 3 a 1 b 2 + a 2 b 1 = 0

Find an equation of a circle through the origin, making an intercept of 10 on the line y = 2 x + 5 2 and subtending an angle of 45 ° at the origin. The centre of the circle is in the positive quadrant.

Let an equation of the circle through the origin be x 2 + y 2 + 2 g x + 2 f y = 0 . . . . i If r is the radius of this circle, then r 2 = g 2 + f 2 . . . . ii Let P Q be the intercept of length 10 made by the line y = 2 x + 5 2 on the circle. Since it is given that P Q subtends an angle of 45 ° at the origin (a point on the circle). It subtends a right angle at the centre - g , - f of the circle. P Q 2 = C P 2 + C Q 2 ⇒ 10 2 = r 2 + r 2 ⇒ r = 5 So, g 2 + f 2 = r 2 = 5 . . . . i Also, C L is perpendicular from C on P Q , so C L = L P = L Q = 10 2 ⇒ 2 g - f - 5 2 1 + 4 = 10 2 = 5 2 ⇒ 2 g - f - 5 2 = ± 5 2 ⇒ 2 g - f = 5 2 ± 5 2 ⇒ 2 g - f = 10 2 = 5 2 or 2 g - f = 0 2 g - f = 5 2 is giving imaginary values, so 2 g - f = 0 ⇒ f = 2 g , solving with i we get g = ± 1 , f = ± 2 Equation of circles are x 2 + y 2 - 2 x - 4 y = 0 or x 2 + y 2 - 2 x + 4 y = 0

If x + y - 1 = tan ⁡ u + v - 1 , where x , y , u and v are all real, prove that the curve u = c o n s t a n t give a family of coaxial circles passing through the points 0 , ± 1 , and that the curves v = c o n s t a n t give a system of circles cutting the first system orthogonally.

Given: tan u + i v = x + i y . . . . i tan u - i v = x - i y . . . . ii Hence, tan u + i v + u - i v = tan u + i v + tan u - i v 1 - tan u + i v tan u - i v ⇒ tan 2 u = x + i y + x - i y 1 - x + i y x - i y ⇒ tan 2 u = 2 x 1 - x 2 - y 2 ⇒ tan 2 u 1 - x 2 - y 2 = 2 x ⇒ x 2 + y 2 + 2 x cot 2 u - 1 = 0 . . . . . ( iii ) Similarly, tan u + i v - u - i v = x + i y - x - i y 1 + x + i y x - i y ⇒ i tan 2 v = 2 i y 1 + x 2 + y 2 ⇒ x 2 + y 2 - 2 y cot 2 v + 1 = 0 . . . . . ( iv ) Solving ( iii ) with y - axis , i.e. x = 0 , we get y 2 - 1 = 0 or y = ± 1 Hence, ( iii ) passes through 0 , ± 1 for all values of u . Clearly ( iii ) pass ( iv ) represent different systems of coaxial circles. Again for ( iii ) and ( iv ) respectively g = cot 2 u , f = 0 , c = - 1 and g ' = 0 , f ' = - cot 2 v , c ' = 1 . Hence, 2 g g ' + 2 f f ' - c - c ' = 0 + 1 - 1 = 0 Therefore, the two systems cut orthogonally.

HARD

JEE Main/Advanced

IMPORTANT

Earn 100

The line $y = x$ touches a circle at $P$ so that $O P = 4 \sqrt{2},$ where $O$ is the origin. The point $(- 10, 2)$ lies inside the circle, and the length of the chord $x + y = 0$ is $6 \sqrt{2} .$ Find the equation of the circle.

Important Questions on Circle

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 4 - Circle>EXERCISE ON LEVEL-II>Q 2

If the points $(α_{1}, β_{1}), (α_{2}, β_{2}), (α_{3}, β_{3})$ lies on the circle $x^{2} + y^{2} + 2 g x + 2 f y = 0$ , then prove that the four straight lines
$α_{1} x + β_{1} y - m (α_{1}^{2} + β_{1}^{2}) = 0$

$α_{2} x + β_{2} y - m (α_{2}^{2} + β_{2}^{2}) = 0$

$α_{3} x + β_{3} y - m (α_{3}^{2} + β_{3}^{2}) = 0, f x - g y = 0$
meet in a point.

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 4 - Circle>EXERCISE ON LEVEL-II>Q 3

Prove that the equation of the circumcircle of the triangle formed by the lines

$u_{1} \equiv a_{1} x + b_{1} y + c_{1} = 0$

$u_{2} \equiv a_{2} x + b_{2} y + c_{2} = 0$

$u_{3} \equiv a_{3} x + b_{3} y + c_{3} = 0$

is $|\begin{matrix} \frac{1}{u_{1}} & \frac{1}{u_{2}} & \frac{1}{u_{3}} \\ a_{2} a_{3} - b_{2} b_{3} & a_{3} a_{1} - b_{3} b_{1} & a_{1} a_{2} - b_{1} b_{2} \\ a_{2} b_{3} + a_{3} b_{2} & a_{3} b_{1} + a_{1} b_{3} & a_{1} b_{2} + a_{2} b_{1} \end{matrix}| = 0$

or $|\begin{matrix} \frac{a_{1}^{2} + b_{1}^{2}}{u_{1}} & \frac{a_{2}^{2} + b_{2}^{2}}{u_{2}} & \frac{a_{3}^{2} + b_{3}^{2}}{u_{3}} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix}| = 0$

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 4 - Circle>EXERCISE ON LEVEL-II>Q 4

Prove that the general equation of circles cutting the circles $x^{2} + y^{2} + 2 g_{i} x + 2 f_{I} y + c_{i} = 0; i = 1, 2$ orthogonally is

$|\begin{array}{l} x^{2} + y^{2} & - x & - y \\ c_{1} & g_{1} & f_{1} \\ c_{2} & g_{2} & f_{2} \end{array}| + k |\begin{matrix} - x & - y & 1 \\ g_{1} & f_{1} & 1 \\ g_{2} & f_{2} & 1 \end{matrix}| = 0$

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 4 - Circle>EXERCISE ON LEVEL-II>Q 5

An isosceles right-angled triangle, whose sides are $1, 1, \sqrt{2}$ lies entirely in the first quadrant with the ends of the hypotenuse on the co-ordinate axes. If it slides, prove that the locus of its centroid is

${(3 x - y)}^{2} + {(x - 3 y)}^{2} = \frac{32}{9}$

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 4 - Circle>EXERCISE ON LEVEL-II>Q 6

Find the locus of the point of intersection of two perpendicular lines each of which touches one of the two circles

{(x - a)}^{2} + y^{2} = b^{2}, {(x + a)}^{2} + y^{2} = c^{2}

and prove that the bisectors of the angles between the straight lines always touch one or the other fixed circles.

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 4 - Circle>EXERCISE ON LEVEL-II>Q 7

Find an equation of the circle which touches the straight lines

x + y = 2, x - y = 2

and also touches the circle

x^{2} + y^{2} = 1

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 4 - Circle>EXERCISE ON LEVEL-II>Q 8

Find an equation of a circle through the origin, making an intercept of

\sqrt{10}

on the line

y = 2 x + \frac{5}{\sqrt{2}}

and subtending an angle of

45 °

at the origin. The centre of the circle is in the positive quadrant.

HARD

JEE Main/Advanced

IMPORTANT

Skills in Mathematics Coordinate Geometry for JEE Main & Advanced>Chapter 4 - Circle>EXERCISE ON LEVEL-II>Q 9

x + y \sqrt{- 1} = \tan (u + v \sqrt{- 1})

, where

x, y, u

and

v

are all real, prove that the curve

u = c o n s t a n t

give a family of coaxial circles passing through the points

(0, \pm 1)

, and that the curves

v = c o n s t a n t

give a system of circles cutting the first system orthogonally.

The line y=x touches a circle at P so that OP=42, where O is the origin. The point (-10,2) lies inside the circle, and the length of the chord x+y=0 is 62. Find the equation of the circle.

Important Questions on Circle

Learn and Prepare for any exam you want !

The line $y = x$ touches a circle at $P$ so that $O P = 4 \sqrt{2},$ where $O$ is the origin. The point $(- 10, 2)$ lies inside the circle, and the length of the chord $x + y = 0$ is $6 \sqrt{2} .$ Find the equation of the circle.