Question

If a circle ${{x}^{2}}+{{y}^{2}}=9$ touches the circle ${{x}^{2}}+{{y}^{2}}+6y+c=0$, then c is equal to
$\begin{align}
  & \text{A}\text{. -27} \\
 & \text{B}\text{. 36} \\
 & \text{C}\text{. -36} \\
 & \text{D}\text{. 27} \\
\end{align}$

Answer 1

Hint: To solve this question first we compare the given equations of the circle with the general equation of the circle. For finding the solution first we need to calculate the radius and center of the circle given in the question. If ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ is the general equation of the circle then, the center of circle is given by $\left( -g,-f \right)$ and radius of circle is given by $r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$ .
Then, by using the relation between the center of the circle and radius of the circle we calculate the value of c.

Complete step-by-step answer:
We have given that a circle ${{x}^{2}}+{{y}^{2}}=9$ touches the circle ${{x}^{2}}+{{y}^{2}}+6y+c=0$.

We have to find the value of $c$ .
Now, we know that if ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ is the general equation of the circle then, the center of circle is given by $\left( -g,-f \right)$ and radius of circle is given by $r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$
Now consider the first circle ${{x}^{2}}+{{y}^{2}}=9$
On comparing with the general equation of circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$, we get
$\begin{align}
  & g=0 \\
 & f=0 \\
 & c=-9 \\
\end{align}$
So, the center of circle will be ${{C}_{1}}=\left( 0,0 \right)$ and the radius will be
 $r=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$
$\begin{align}
  & {{r}_{1}}=\sqrt{{{0}^{2}}+{{0}^{2}}+9} \\
 & {{r}_{1}}=\sqrt{9} \\
 & {{r}_{1}}=3 \\
\end{align}$
Now, let us consider circle ${{x}^{2}}+{{y}^{2}}+6y+c=0$
On comparing with the general equation of circle ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$, we get
$\begin{align}
  & g=0 \\
 & f=3 \\
 & c=c \\
\end{align}$
So, the center of circle will be ${{C}_{2}}=\left( 0,-3 \right)$ and the radius will be
 ${{r}_{2}}=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$
$\begin{align}
  & {{r}_{2}}=\sqrt{{{0}^{2}}+{{3}^{2}}-c} \\
 & {{r}_{2}}=\sqrt{9-c} \\
\end{align}$
Now, we know that if two circles touch each other then, the relation between their centers and radius is given by
${{C}_{1}}{{C}_{2}}={{r}_{2}}-{{r}_{1}}$
Substituting the values, we get
$\begin{align}
  & \Rightarrow 3=\sqrt{9-c}-3 \\
 & \Rightarrow 3+3=\sqrt{9-c} \\
 & \Rightarrow 6=\sqrt{9-c} \\
 & \Rightarrow 36=9-c \\
 & \Rightarrow 36-9=-c \\
 & -c=27 \\
 & c=-27 \\
\end{align}$
The value of c is $-27$ .

So, the correct answer is “Option A”.

Note: The relation between the center and radius of two circles touch each other is also given by ${{C}_{1}}{{C}_{2}}={{r}_{1}}-{{r}_{2}}$ and ${{C}_{1}}{{C}_{2}}={{r}_{1}}+{{r}_{2}}$. But when we put the values and solve these equations we get the answer $c=9$, which is not matches the options given so we use the relation ${{C}_{1}}{{C}_{2}}={{r}_{2}}-{{r}_{1}}$.
$\begin{align}
  & {{C}_{1}}{{C}_{2}}={{r}_{1}}-{{r}_{2}} \\
 & 3=3-\sqrt{9-c} \\
 & 3-3=-\sqrt{9-c} \\
 & -\sqrt{9-c}=0 \\
 & c=9 \\
\end{align}$
And
\[\begin{align}
  & {{C}_{1}}{{C}_{2}}={{r}_{1}}+{{r}_{2}} \\
 & 3=3+\sqrt{9-c} \\
 & 3-3=\sqrt{9-c} \\
 & 0=\sqrt{9-c} \\
 & c=9 \\
\end{align}\]