Question

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

• A. $(3/2,1/2)$
• B. $(1/2,3/2)$
• C. $(1/2,1/2)$
• D. $(1/2,12^{1/2})$

Answer 1

Answer: (D)

Let $C_1(h,k)$ be the centre of the required circle. Then,
$\sqrt{(h-0{)}^2+(k-0{)}^2}=\sqrt{(h-1{)}^2+(k-0{)}^2}$
$\Rightarrow h^2+k^2=h^2-2h+1+k^2$
$\Rightarrow -2h+1=0\Rightarrow h=1/2$
Since, $(0,0)$ and $(1,0)$ lie inside the circle $x^2+y^2=9$.
Therefore, the required circle can touch the given circle
internally.
i.e. $C_1.C_2=r_1\sim r_2$
$\Rightarrow \sqrt{h^2+k^2}=3-\sqrt{h^2+K^2}$
$\Rightarrow 2\sqrt{h^2+k^2}=3\Rightarrow 2\sqrt{\frac{1}{4}+K^2=3}$
$\Rightarrow \sqrt{\frac{1}{4}+k^2}=\frac{3}{2}\Rightarrow \frac{1}{4}+k^2=\frac{9}{4}$
$\Rightarrow k^2=2\Rightarrow k=\pm \sqrt{2}$