A circle C passes through (2a, 0) and the lin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$. Let the equation of circle $C$ be $x^2 + y^2 + 2gx + 2fy + c

Vedclass · Accepted Answer

centre of $C$ is $(a, 0)$ and $C$ passes through $(0, 0)$ and $(a, a)$

Vedclass · Answer

(C) The radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$. Let the equation of circle $C$ be $x^2 + y^2 + 2gx + 2fy + c = 0$. The radical axis of $C$ and $x^2 + y^2 - a^2 = 0$ is $(x^2 + y^2 + 2gx + 2fy + c) - (x^2 + y^2 - a^2) = 0$,which simplifies to $2gx + 2fy + c + a^2 = 0$. Given the radical axis is $2x = a$ or $x - a/2 = 0$. Comparing $2gx + 2fy + c + a^2 = 0$ with $x - a/2 = 0$,we get $2g = k$,$2f = 0$,and $c + a^2 = -ak/2$ for some constant $k$. Since $2f = 0$,$f = 0$. Thus,the circle $C$ is $x^2 + y^2 + 2gx + c = 0$. The radical axis is $2gx + c + a^2 = 0$. Comparing with $x = a/2$,we get $2g = 1$ and $c + a^2 = -a/2$. Wait,using the family of circles method: The equation of any circle passing through the intersection of $x^2 + y^2 - a^2 = 0$ and $x - a/2 = 0$ is $(x^2 + y^2 - a^2) + \lambda(x - a/2) = 0$. Since it passes through $(2a, 0)$,we have $(4a^2 - a^2) + \lambda(2a - a/2) = 0$,which gives $3a^2 + \lambda(3a/2) = 0$,so $\lambda = -2a$. The equation of circle $C$ is $x^2 + y^2 - a^2 - 2a(x - a/2) = 0$,which simplifies to $x^2 + y^2 - 2ax = 0$. This is $(x - a)^2 + y^2 = a^2$. The centre is $(a, 0)$. It passes through $(0, 0)$ and $(a, a)$ (since $(a-a)^2 + a^2 = a^2$).

$A$ circle $C$ passes through $(2a, 0)$ and the line $2x = a$ is the radical axis of the circle $C$ and the circle $x^2 + y^2 = a^2$. Then,

Similar Questions

The length of the tangent drawn from any point on the circle $x^2+y^2+2gx+2fy+c_1=0$ to the circle $x^2+y^2+2gx+2fy+c_2=0$ is

The number of common tangents to the circles ${x^2} + {y^2} - 4x - 6y - 12 = 0$ and ${x^2} + {y^2} + 6x + 18y + 26 = 0$ is

The equation of the circle which passes through the centre of the circle $x^2+y^2+8x+10y-7=0$ and is concentric with the circle $2x^2+2y^2-8x-12y-9=0$ is

The number of common tangents to the circles $x^2+y^2+4x=0$ and $x^2+y^2-2x=0$ is

$A$ circle $C$ touches the line $x=2y$ at the point $(2,1)$ and intersects the circle $C_{1}: x^{2}+y^{2}+2y-5=0$ at two points $P$ and $Q$ such that $PQ$ is a diameter of $C_{1}$. Then the diameter of $C$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module