Question

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

• A. centre of $C$ is $(-a,0)$ and $C$ passes through $(0,0)$ and $(-a,-a)$
• B. circle $C$ is $x^2+y^2-2ax-2ay=0$
• C. centre of $C$ is $(a,0)$ and $C$ passes through $(0,0)$ and $(a,a)$
• D. centre of $C$ is $(0,-a)$ and $C$ passes through $(-a,-a)$ and $(0,0)$

Answer 1

Answer: (C)

Radical axis $PQ$ is $x-\frac{a}{2}=0$ which cuts the circle $x^2+y^2=a^2$ in $P$ and $Q$. Now, any circle which passes through the points $P$ and $Q$ will have radical axis the line $PQ$ with respect to $x^2+y^2-a^2=0.$
Hence, its equation is the equation of the circle through the points of intersection of the circle.
$x^2+y^2-a^2=0$
and the line $x-\frac{a}{2}=0$ and is given by $s+\lambda p=0$
i.e. $\left(x^2+y^2-a^2\right)+\lambda \left(x-\frac{a}{2}\right)=0$
As, it passes through the point $(2a,0)$.
$\therefore \left(4a^2-a^2\right)+\lambda \left(2a-\frac{a}{2}\right)=0$
$\Rightarrow \lambda =-2a$
Hence, the equation of circle $C$ is
$\left(x^2+y^2-a^2\right)-2a\left(x-\frac{a}{2}\right)=0$
$\Rightarrow x^2+y^2-2ax=0$
$\Rightarrow (x-a{)}^2+y^2=a^2$
Whose centre is $(a,0)$ and passes through $(0,0)$ and $(a,a)$.