Question

Given the three circles ${{x}^{2}}+{{y}^{2}}-16x+60=0$, $3{{x}^{2}}+3{{y}^{2}}-36x+81=0$, and ${{x}^{2}}+{{y}^{2}}-16x-12y+84=0$, find (1) the point from which the tangents to them are equal in length, and (2) this length.

Answer 1

Hint: Use the fact that the length of the tangent from any point $\left( {{x}_{0}},{{y}_{0}} \right)$ to a given circle having equation S can be found out by substituting the point in the equation and taking its square root. Thus, assume a general point $\left( h,k \right)$ from which the tangents are equal in length and substitute this in the equation of the three circles to find the values of h and k. The point $\left( h,k \right)$ so obtained will give us the point from which the tangents to the given circles are equal in length.

Complete step-by-step answer:

Let us assume that the required point is given by $\left( h,k \right)$. Let us further denote the three circles as
$\begin{align}
  & {{C}_{1}}:{{x}^{2}}+{{y}^{2}}-16x+60=0 \\
 & {{C}_{2}}:3{{x}^{2}}+3{{y}^{2}}-36x+81=0 \\
 & {{C}_{3}}:{{x}^{2}}+{{y}^{2}}-16x-12y+84=0 \\
\end{align}$
The equation of ${{C}_{2}}$ can be re-written by dividing both sides by 3 to give us
${{C}_{2}}:{{x}^{2}}+{{y}^{2}}-12x+27=0$
Now, we calculate the length of the tangents from the point $\left( h,k \right)$ to these three circles by substituting this point in the three equations and taking the square root. Let us further denote these three lengths as ${{l}_{1}},{{l}_{2}},{{l}_{3}}$. Thus, we get the three lengths as
$\begin{align}
  & {{l}_{1}}=\sqrt{{{h}^{2}}+{{k}^{2}}-16h+60} \\
 & {{l}_{2}}=\sqrt{{{h}^{2}}+{{k}^{2}}-12h+27} \\
 & {{l}_{3}}=\sqrt{{{h}^{2}}+{{k}^{2}}-16h-12k+84} \\
\end{align}$
We now equate these three lengths to get the values of h and k.
Equating ${{l}_{1}}$ and ${{l}_{2}}$, we get
$\sqrt{{{h}^{2}}+{{k}^{2}}-16h+60}=\sqrt{{{h}^{2}}+{{k}^{2}}-12h+27}$
Squaring both sides of this equation, we get
$\begin{align}
  &{{{h}^{2}}}+{{{k}^{2}}}-16h+60={{{h}^{2}}}+{{{k}^{2}}}-12h+27 \\
 & \Rightarrow -16h+60=-12h+27 \\
 & \Rightarrow 60-27=16h-12h \\
 & \Rightarrow 33=4h \\
 & \Rightarrow h=\dfrac{33}{4} \\
\end{align}$
Similarly, equating ${{l}_{1}}$ and ${{l}_{3}}$, we get
$\sqrt{{{h}^{2}}+{{k}^{2}}-16h+60}=\sqrt{{{h}^{2}}+{{k}^{2}}-16h-12k+84}$
Squaring both sides of this equation, we get
\[\begin{align}
  & {{h}^{2}}+{{k}^{2}}-16h+60={{h}^{2}}+{{k}^{2}}-16h-12k+84 \\
 & \Rightarrow 60=-12k+84 \\
 & \Rightarrow 12k=84-60 \\
 & \Rightarrow k=\dfrac{24}{12} \\
 & \Rightarrow k=2 \\
\end{align}\]
Thus, the values of h and k come out to be $\dfrac{33}{4}$ and 2 respectively to give the point $\left( \dfrac{33}{4},2 \right)$ as the point from which the three tangents drawn to circles ${{C}_{1}},{{C}_{2}}$ and ${{C}_{3}}$ are equal in length. To find this length, we can calculate ${{l}_{1}}$ as
$\begin{align}
  & {{l}_{1}}=\sqrt{{{\left( \dfrac{33}{4} \right)}^{2}}+{{\left( 2 \right)}^{2}}-16\left( \dfrac{33}{4} \right)+60} \\
 & \Rightarrow {{l}_{1}}=\sqrt{\left( \dfrac{1089}{16} \right)+4-132+60} \\
 & \Rightarrow {{l}_{1}}=\sqrt{\dfrac{1089}{16}-68} \\
 & \Rightarrow {{l}_{1}}=\sqrt{\dfrac{1089-1088}{16}} \\
\end{align}$
$\begin{align}
  & \Rightarrow {{l}_{1}}=\sqrt{\dfrac{1}{16}} \\
 & \Rightarrow {{l}_{1}}=\dfrac{1}{4} \\
\end{align}$
Thus, the length of these three tangents is $\dfrac{1}{4}$ units.

Note: The division by 3 to get a reduced equation of the circle ${{C}_{2}}$ is important to keep in mind as otherwise, in the calculation, the square terms don’t cancel out and we get a quadratic equation which might be more difficult to solve and make the solution lengthier. It will also increase the chances of error. Also, there is only one point that we have obtained in this case because there are three circles. Had there been only two circles, we would have obtained a line and not a point, and that line is known as the radical axis of the two circles.