Question

The centers of those circles which touches the circle, ${{x}^{2}}+{{y}^{2}}-8x-8y-4=0$ externally and also touch the $x$-axis, lie on ?

Answer 1

Hint: For these kinds of questions, we assume the coordinates of the center of the circle which is touching the other circle and the axis. And using the given conditions, we build up our own equations. At the end , after we get our final equation, we substitute the variables in the equation we get with $x,y$ so that the general equation of a conic section or a line or any curve is obtained. So, let us the equation of the circle which is touching the other circle and the axis to be ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ .

Complete step by step solution:
The equation of the circle we assumed is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$. So the center of this circle is $C\left( -g,-f \right)$ .
It is mentioned in the question that the circle touches $x$-axis.
Generally, when the circle lies above the $x$-axis, the condition is $\sqrt{{{g}^{2}}-c}\ge 0$ .
Since it is touching $x$-axis, the condition becomes $\sqrt{{{g}^{2}}-c}=0$ .
From this, we infer the following :
$\begin{align}
  & \Rightarrow \sqrt{{{g}^{2}}-c}=0 \\
 & \Rightarrow {{g}^{2}}=c \\
\end{align}$
Let us substitute in the equation of the circle.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow {{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}+2gx+2fy+{{g}^{2}}=0 \\
\end{align}$
The other condition is that this circle and the circle ${{x}^{2}}+{{y}^{2}}-8x-8y-4=0$are touching externally.
The condition for two circles with centers ${{C}_{1}},{{C}_{2}}$ and radii ${{r}_{1}},{{r}_{2}}$ respectively to touch externally is ${{C}_{1}}{{C}_{2}}={{r}_{1}}+{{r}_{2}}$ .
The center of the circle ${{x}^{2}}+{{y}^{2}}-8x-8y-4=0$ is ${{C}_{2}}\left( 4,4 \right)$ and ${{r}_{2}}=\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{{{4}^{2}}+{{4}^{2}}+4}=\sqrt{36}=6$ .
Here our ${{C}_{1}}$ is $C\left( -g,-f \right)$and our ${{r}_{1}}=\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{{{g}^{2}}+{{f}^{2}}-{{g}^{2}}}=f$ .
Let us apply the condition.
Upon substituting in the given condition, we get the following :
$\begin{align}
  & \Rightarrow {{C}_{1}}{{C}_{2}}={{r}_{1}}+{{r}_{2}} \\
 & \Rightarrow \sqrt{{{\left( g+4 \right)}^{2}}+{{\left( f+4 \right)}^{2}}}=6+f \\
\end{align}$
Let us square it on both sides.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow {{C}_{1}}{{C}_{2}}={{r}_{1}}+{{r}_{2}} \\
 & \Rightarrow \sqrt{{{\left( g+4 \right)}^{2}}+{{\left( f+4 \right)}^{2}}}=6+f \\
 & \Rightarrow {{g}^{2}}+16+8g+{{f}^{2}}+8f+16={{f}^{2}}+36+12f \\
 & \Rightarrow {{g}^{2}}+8g=4f+4 \\
\end{align}$
Now let us substitute $g,f$ with $x,y$ so as to check which conic or line or curve we get.
Upon doing so, we get the following :
$\begin{align}
  & \Rightarrow {{g}^{2}}+8g=4f+4 \\
 & \Rightarrow {{x}^{2}}+8x=4y+4 \\
\end{align}$
As we can see clearly, it is a parabola.
$\therefore $ The centers of those circles which touches the circle,${{x}^{2}}+{{y}^{2}}-8x-8y-4=0$ externally and also touch the $x$-axis, lie on a parabola. So the option is D.

Note: We can either complete the squares of the last equation we got to conclude that it is a parabola or if we are well versed with all the conic sections and their general equation, we can directly conclude that it is a parabola. So we should know the general equations of all the conic sections so as to complete a question quickly in the exam. We should also remember the conditions for different positions of the circles with the axes as well as with other circles.