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Q.
The centre of the circle passing through $(0,0)$ and $(1,0)$ and touching the circle $ x^2 + y^2 =9 $ is
ManipalManipal 2015
Solution:
Let the equation of the circle be
$x^{2}+y^{2}+2 g x+2 f y+c=0$
which passes through $(0,0)$ and $(1,0)$.
$\therefore c=0$ and $1+2 g+c=0$
$\Rightarrow c=0, g=-\frac{1}{2}$
The circle $x^{2}+y^{2}+2 g x+2 f y+c=0$
touches the circle $x^{2}+y^{2}=9$.
$ \therefore \sqrt{g^{2}+f^{2}} =3 \pm \sqrt{g^{2}+f^{2}-c} $
$\Rightarrow \sqrt{\frac{1}{4}+f^{2}} =3 \pm \sqrt{\frac{1}{4}+f^{2}}$
$\Rightarrow 3 =2 \sqrt{\frac{1}{4}+f^{2}} $
$\Rightarrow f^{2} =\frac{9}{4}-\frac{1}{4}=2$
$\Rightarrow f =\pm \sqrt{2}$
Hence, the coordinates of the centre are
$\left(\frac{1}{2}, \pm \sqrt{2}\right).$