Question

A circle passes through the points $(0,0)$ and $(0,1)$ also touches the circle $x^2+y^2=16$.

The radius of the circle is

• A. $1$
• B. $2$
• C. $3$
• D. $4$
• E. $5$

Answer 1

The correct option is B

$2$

Explanation for correct option:

Step 1. Assume the general equation of the circle :

Let the equation of the circle be

$x^2+y^2+2gx+2fy+c=0$

Step 2. Apply the given condition :

The circle $x^2+y^2+2gx+2fy+c=0$ passes through the $(0,0)$ and $(0,1)$.

$\therefore c=0$ and $1+2f+c=0$

$\Rightarrow c=0,f=-\frac{1}{2}$

The circle $x^2+y^2+2gx+2fy+c=0$ touches the circle $x^2+y^2=16$

$$\begin{gathered} \therefore \sqrt{g^2+f^2}=4\pm \sqrt{g^2+f^2-c} \\ \Rightarrow \sqrt{g^2+\frac{1}{4}}=4\pm \sqrt{g^2+\frac{1}{4}} \\ \Rightarrow 4=2\sqrt{g^2+\frac{1}{4}} \\ \Rightarrow g^2=4-\frac{1}{4} \\ \Rightarrow g^2=\frac{15}{4} \\ \Rightarrow g=\pm \frac{\sqrt{15}}{2} \end{gathered}$$

Step 3. Calculate the center and radius of the circle :

Therefore, the coordinate of the center is $=\left(\pm \frac{\sqrt{15}}{2},\frac{1}{2}\right)$

And radius $\sqrt{g^2+f^2-c}=\sqrt{\frac{15}{4}+\frac{1}{4}}$

$$\begin{gathered} =\sqrt{\frac{16}{4}} \\ =\sqrt{4} \\ =\mathbf{2} \end{gathered}$$

Hence, the correct option is (B).