Question

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

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

Answer 1

The correct option is D

$\left(\frac{1}{2},-2^{1/2}\right)$

Explanation for the correct option:

Finding the center of the circle:

The general equation of the circle is $x^2+y^2+2gx+2fy+c=0$ with center $\left(-g,-f\right)$ and radius $\sqrt{g^2+f^2-c}$

Circle passes through the point $\left(0,0\right)$.

So, substitute $x=0$ and $y=0$ in $x^2+y^2+2gx+2fy+c=0$.

$\Rightarrow c=0$

Similarly, Circle passes through the point $\left(1,0\right)$.

So, substitute $x=1$ and $y=0$ in $x^2+y^2+2gx+2fy=0$

.

$$\begin{gathered} \Rightarrow 1+2g=0 \\ \Rightarrow g=-1/2 \end{gathered}$$

Now the circle touches another circle with equation $x^2+y^2=9$.

Since the center of the this circle lies on the circle $1$, the radius of a new circle is equal to the diameter of the circle $1$.

$$\begin{gathered} \therefore 2\sqrt{g^2+f^2-c}=3 \\ \Rightarrow \sqrt{{\left(\frac{1}{4}\right)}^2+f^2}=\frac{3}{2} \\ \Rightarrow {\left(\frac{1}{2}\right)}^2+f^2=\frac{9}{4} \\ \Rightarrow f^2=\frac{9}{4}-\frac{1}{4} \\ =2 \\ \Rightarrow f=\pm \sqrt{2} \end{gathered}$$

Hence the centres of circle $1$ is $\left(-g,-f\right)=\left(\frac{1}{2},\pm \sqrt{2}\right)$ .

Therefore, the correct answer is option (D).