Question

Find the radical centre of the three circles ${{x}^{2}}+{{y}^{2}}={{a}^{2}},{{\left( x-c \right)}^{2}}+{{y}^{2}}={{a}^{2}}\ $and ${{x}^{2}}+{{\left( y-b \right)}^{2}}={{a}^{2}}$.

Answer 1

- Hint: We will be using the concept of circles and coordinate geometry to solve problems. Also, we will be using the concept of radical axes and radical centre.

Complete step-by-step solution -

We have to find the radical centre of three circles given to us.
Let,
$\begin{align}
  & {{S}_{1}}={{x}^{2}}+{{y}^{2}}={{a}^{2}} \\
 & {{S}_{2}}={{\left( x-c \right)}^{2}}+{{y}^{2}}={{a}^{2}}\ \\
 & {{S}_{3}}={{x}^{2}}+{{\left( y-b \right)}^{2}}={{a}^{2}} \\
\end{align}$
We know that the radical centre is the point of intersection of radical axes of the pair of circles. We also know that if ${{S}_{1}},{{S}_{2}},{{S}_{3}}$ are the pair of circles then their radical axes are given by ${{S}_{1}}-{{S}_{2}}=0,{{S}_{2}}-{{S}_{3}}=0,{{S}_{3}}-{{S}_{1}}=0$
So,
$\begin{align}
  & {{S}_{1}}-{{S}_{2}}={{x}^{2}}+{{y}^{2}}-{{a}^{2}}-{{\left( x-c \right)}^{2}}-{{y}^{2}}+{{a}^{2}}=0 \\
 & ={{x}^{2}}-{{\left( x-c \right)}^{2}}=0 \\
 & \text{Using }\ {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) \\
 & \left( x-x+c \right)\left( 2x-c \right)=0 \\
 & c\left( 2x-c \right)=0 \\
 & x=\dfrac{c}{2}............\left( 1 \right) \\
\end{align}$
Now,
$\begin{align}
  & {{S}_{2}}-{{S}_{3}}={{\left( x-c \right)}^{2}}+{{y}^{2}}-{{a}^{2}}-{{x}^{2}}-{{\left( y-b \right)}^{2}}+{{a}^{2}} \\
 & ={{\left( x-c \right)}^{2}}-{{x}^{2}}+{{y}^{2}}-{{\left( y-b \right)}^{2}} \\
 & =\left( 2x-c \right)\left( -c \right)+b\left( 2y-b \right) \\
 & \left( 2x-c \right)\left( -c \right)+b\left( 2y-b \right)=0 \\
\end{align}$
Using equation (1), we get,
$\begin{align}
  & \left( 2\left( \dfrac{c}{2} \right)-c \right)\left( -c \right)+b\left( 2y-b \right)=0 \\
 & 0\left( -c \right)+b\left( 2y-b \right)=0 \\
 & 2y-b=0 \\
 & y=\dfrac{b}{2}........\left( 2 \right) \\
\end{align}$
Thus, from equation (1) and (2) we get radical centre $\left( \dfrac{c}{2},\dfrac{b}{2} \right)$.

Note: To solve these types of questions it is important to remember the concept of radical centre and radical axes. It should be noted that the radical centre is the point of intersection of the three radical axes also to find the radical axes of two circle we subtract the one equation from the other for example if
$\begin{align}
  & {{S}_{1}}={{x}^{2}}+{{y}^{2}}={{a}^{2}} \\
 & {{S}_{2}}={{\left( x-c \right)}^{2}}+{{y}^{2}}={{a}^{2}}\ \\
 & {{S}_{3}}={{x}^{2}}+{{\left( y-b \right)}^{2}}={{a}^{2}} \\
\end{align}$
then the equations of radical axes are ${{S}_{1}}-{{S}_{2}}=0,{{S}_{2}}-{{S}_{3}}=0,{{S}_{3}}-{{S}_{1}}=0$ .