Question

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

• A. $(\frac{3}{2},\frac{1}{2})$
• B. $(\frac{1}{2},\frac{3}{2})$
• C. $(\frac{1}{2},2^{\frac{1}{2}})$
• D. $(\frac{1}{2},{-2}^{\frac{1}{2}})$

Answer 1

Answer: (C), (D)

The correct options are
C $(\frac{1}{2},2^{\frac{1}{2}})$
D $(\frac{1}{2},{-2}^{\frac{1}{2}})$

let the equation of the circle passing through the origin be $x^2+y^2-2hx-2ky=0$

Since, it passes through $(1,0)$

then $1=2h$

$\Rightarrow h=\frac{1}{2}$

Since it touches the circle $x^2+y^2=9$

then $C_1{C}_2=r_1-r_2$

$\Rightarrow \sqrt{h^2+k^2}=3-\sqrt{h^2+k^2}$

$\Rightarrow k=\pm \sqrt{2}$