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Q.
A circle $C$ passes through $(2 a, 0)$ and the line $2 x=a$ is the radical axis of the circle $C$ and the circle $x^{2}+y^{2}=a^{2}$, then
TS EAMCET 2018
Solution:
Radical axis $P Q$ 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 $P Q$ 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 $(2 a, 0)$.
$\therefore \left(4 a^{2}-a^{2}\right)+\lambda\left(2 a-\frac{a}{2}\right)=0$
$\Rightarrow \lambda=-2 a$
Hence, the equation of circle $C$ is
$\left(x^{2}+y^{2}-a^{2}\right)-2 a\left(x-\frac{a}{2}\right)=0$
$\Rightarrow x^{2}+y^{2}-2 a x=0$
$\Rightarrow (x-a)^{2}+y^{2}=a^{2}$
Whose centre is $(a, 0)$ and passes through $(0,0)$ and $(a, a)$.