Question

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

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

Answer 1

Answer: (B)

Let equation of circle be $x^2+y^2+2gx+2fy+c=0$
As it passes through $(0,0)$ so $c=0$

and as it passes through $(1,0)$ so$-g=\frac{1}{2}$

Now $x^2+y^2+2gx+2fy+c=0$ and $x^2+y^2=9$ touches each

other

$\therefore$ Equation of common tangent is $2gx+2fy-9=0$ and distance from the centre of circle $x^2+y^2=9$ to the.common tangent is equal to the radius of the, circle $x^2+y^2=9$

$\therefore \frac{|0+0-9|}{\sqrt{4g^2+4f^2}}=3$

$\Rightarrow 3^2=4(g^2+f^2)$

$9=4(\frac{1}{4}+f^2)$ $9=1+4f^2$

$\therefore f^2=2$

$\Rightarrow f=\pm \sqrt{2}\therefore -f=\pm \sqrt{2}$

$\therefore$ Centre of the required circle be $(\frac{1}{2},\sqrt{2}),(\frac{1}{2},-\sqrt{2}),$