Question

The centre of the circle which passes through $(0,0),(1,0)$, and touches the circle $x^2+y^2=9$ is

• A. $(-2,\sqrt{2})$
• B. $(\frac{1}{2},-2)$
• C. $(\frac{1}{2},-\sqrt{2})$
• D. $(\frac{1}{2},2)$

Answer 1

Answer: (C)

$(\frac{1}{2},-\sqrt{2})$

Let $x^2+y^2+2gx+2fy=0$. Equation of circle passes through $(0,0)$
It also pass through $(1,0)$ so
$1+2g=0$
$g=-\frac{1}{2}$
And, It also touches $x^2+y^2=9$ internally
$r_2-r_1=c_1{c}_2$
$3-\sqrt{g^2+f^2}=\sqrt{g^2+f^2}$
$\sqrt{g^2+f^2}=\frac{3}{2}$
$g^2+f^2=\frac{9}{4}$
$f^2=\frac{9-1}{4}=2$
$f=\sqrt{2},-\sqrt{2}$
So, centre of circle is $(-g,-f)$
$\Rightarrow (\frac{1}{2},-\sqrt{2})or(\frac{1}{2},\sqrt{2})$