Question

Find the equation of the circle with radius 5 whose centre lies on the $x$- axis and passes through the point (2, 3).

Answer 1

Hint: Here, we have been given the radius, the center lies on $x$- axis and passes through a point. First using the equation of circle, ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$ and considering the center of the circle as $\left( h,0 \right)$, we will form an equation and find the values of $h$. Then, find the equation by using the values of $h$.

Complete step-by-step answer:
The given conditions for the equation of the circle is that we have a radius, r = 5, whose center lies on $x$- axis and the circle also passes through point (2, 3).
Since, the center lies on $x$- axis, we can consider our center to be $\left( h,k \right)=\left( h,0 \right)$
We know, the equation of circle
${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{r}^{2}}$

Let us substitute $\left( h,k \right)$ in the above equation, we get

${{\left( x-h \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\left( 5 \right)}^{2}}$................... (i)
According to the conditions, we know that the circle also passes through the point (2, 3)
So, we can substitute the point (2, 3) in the equation (i), we get
${{\left( 2-h \right)}^{2}}+{{\left( 3 \right)}^{2}}=25$
Let us further expand the expression and find the value of $h$.
$\left( 4-4h+{{h}^{2}} \right)+9=25$
Subtract by – 9 on both the sides of equation, we get
$\begin{align}
  & 4-4h+{{h}^{2}}+9-9=25-9 \\
 & 4-4h+{{h}^{2}}=16
\end{align}$
Now, let us subtract by – 4 on both the sides of the equation, we get
$\begin{align}
  & 4-4h+{{h}^{2}}-4=16-4 \\
 & -4h+{{h}^{2}}=12 \\
 & {{h}^{2}}-4h-12=0
\end{align}$
The above equation which we obtained is in the form of a quadratic equation, let us find the value of $h$ by solving the quadratic equation.
$\begin{align}
  & {{h}^{2}}-4h-12=0 \\
 & {{h}^{2}}-6h+2h-12=0 \\
 & h\left( h-6 \right)+2\left( h-6 \right)=0 \\
 & \left( h+2 \right)\left( h-6 \right)=0 \\
\end{align}$
So, the values of $h$ are – 2 and 6.
Let us find the equation of the circle for both centers (-2, 0) and (6, 0) by substituting the values in equation (i).
When, we have $h=-2$ that is when the center is (-2, 0), we will get
$\begin{align}
  & {{\left[ x-\left( -2 \right) \right]}^{2}}+{{\left( y \right)}^{2}}=25 \\
 & {{\left( x+2 \right)}^{2}}+{{y}^{2}}=25 \\
 & {{x}^{2}}+4x+4+{{y}^{2}}=25 \\
\end{align}$
Let us subtract by 25 on both the sides of the equation, we get
$\begin{align}
  & {{x}^{2}}+4x+4+{{y}^{2}}-25=25-25 \\
 & {{x}^{2}}+{{y}^{2}}+4x-21=0 \\
\end{align}$
We will get a circle just like this with a center of (-2, 0) of radius 5.

When, we have $h=6$ that is when the center is (6, 0), we get
$\begin{align}
  & {{\left( x-6 \right)}^{2}}+{{\left( y \right)}^{2}}=25 \\
 & {{x}^{2}}-12x+36+{{y}^{2}}=25 \\
\end{align}$
Let us subtract by 25 on both the sides of the equation, we get
$\begin{align}
  & {{x}^{2}}-12x+36+{{y}^{2}}-25=25-25 \\
 & {{x}^{2}}+{{y}^{2}}-12x+11=0 \\
\end{align}$
Below is the representation of the circle with a center (6, 0) with a radius of 5.

Hence, the equation of the circle is either ${{x}^{2}}+{{y}^{2}}+4x-21=0$ or ${{x}^{2}}+{{y}^{2}}-12x+11=0$.

Note: In the given question, we have to find the equation of the circle for which the general form of the equation of circle is important. Also, when the point lies on the circumference of the circle it means that it is one of the roots to solve the equation of the circle.