Question

Find the equation of the circle passing through the three points (1,2), (3,-4), (5,-6).

Answer 1

Hint: We need to find the equation of the circle passing through the three given points. We start to solve the problem by substituting the three given points in the general equation of the circle. Then, we need to simplify the equations to get the equation of a circle.

Complete step by step solution:
Let the three given points be $A,B,C\;$ respectively.
$A=\left( 1,2 \right)$
$B=\left( 3,-4 \right)$
$C=\left( 5,-6 \right)$
The general form of the equation of the circle is ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ where $\left( -g,-f \right)$ is the centre of the circle.
We need to substitute the given points in the place of $\left( x,y \right)$ in the equation of a circle.
Substituting the point $A$ in the equation of a circle, we get,
$\Rightarrow {{1}^{2}}+{{2}^{2}}+2g\left( 1 \right)+2f\left( 2 \right)+c=0$
Simplifying the above equation, we get,
$\Rightarrow 1+4+2g+4f+c=0$
Rearranging the terms, we get,
$\Rightarrow 2g+4f+c=-5......\left( 1 \right)$
Substituting the point $B$ in the equation of a circle, we get,
$\Rightarrow {{3}^{2}}+{{\left( -4 \right)}^{2}}+2g\left( 3 \right)+2f\left( -4 \right)+c=0$
Simplifying the above equation, we get,
$\Rightarrow 9+16+6g-8f+c=0$
Rearranging the terms, we get,
$\Rightarrow 6g-8f+c=-25......\left( 2 \right)$
Substituting the point $C$ in the equation of a circle, we get,
$\Rightarrow {{5}^{2}}+{{\left( -6 \right)}^{2}}+2g\left( 5 \right)+2f\left( -6 \right)+c=0$
Simplifying the above equation, we get,
$\Rightarrow 25+36+10g-12f+c=0$
Rearranging the terms, we get,
$\Rightarrow 10g-12f+c=-61......\left( 3 \right)$
Multiplying the equation $\left( 1 \right)$ by 2 and adding with the equation $\left( 2 \right)$ , we get,
$\begin{align}
  & \Rightarrow 4g+8f+2c=-10 \\
 & +\text{ }6g-8f+c=-25 \\
 & - \\
 & \text{ 10}g+3c=-35..................\left( 4 \right) \\
 & - \\
\end{align}$
Multiplying the equation $\left( 1 \right)$ by 3 and adding with the equation $\left( 3 \right)$ , we get,
$\begin{align}
  & \Rightarrow 6g+12f+3c=-15 \\
 & +\text{ 10}g-12f+c=-61 \\
 & - \\
 & \text{ 16}g+4c=-76...................\left( 5 \right) \\
 & - \\
\end{align}$
Multiplying the equation $\left( 4 \right)$ with 3,
$\Rightarrow \left( 10g+3c=-35 \right)\times 4$
$\Rightarrow 40g+12c=-140$
Multiplying the equation $\left( 5 \right)$ with 4,
$\Rightarrow \left( 16g+4c=-76 \right)\times 3$
$\Rightarrow 48g+12c=-228$
Subtracting the above two equations, we get,
$\begin{align}
  & \Rightarrow 40g+12c=-140 \\
 & -\text{ }48g+12c=-228 \\
 & - \\
 & \text{ }-\text{8}g=88 \\
 & - \\
\end{align}$$\Rightarrow g=-\dfrac{88}{8}$
$\therefore g=-11$
Substituting the value of $g$ in the equation $\left( 5 \right)$ , we get,
$\Rightarrow \text{16}\left( -11 \right)+4c=-76$
$\Rightarrow -176+4c=-76$
Shifting $-176$ to the other side of the equation, we get,
$\Rightarrow 4c=-76+176$
$\Rightarrow 4c=100$
$\Rightarrow c=\dfrac{100}{4}$
$\therefore c=25$
Substituting the value of $g,f,c\;$ in the equation $\left( 1 \right)$ , we get,
$\Rightarrow 2\left( -11 \right)+4f+25=-5$
$\Rightarrow -22+4f+25=-5$
$\Rightarrow 4f+3=-5$
$\Rightarrow 4f=-5-3$
$\Rightarrow 4f=-8$
$\Rightarrow f=\dfrac{-8}{4}$
$\therefore f=-2$
$\therefore$ The equation of the circle is ${{x}^{2}}+{{y}^{2}}-22x-4y+25=0$ .
The equation of the circle can be diagrammatically represented as follows,

Here, O is the center of the circle.

Note: The result of the given question can be verified by substituting any of the given points in the question. The result attained is correct only if the left-hand side of the equation is equal to the right-hand side of the equation.
Substituting point $A\left( 1,2 \right)$ in the equation, we get,
$\Rightarrow {{1}^{2}}+{{2}^{2}}-22\left( 1 \right)-4\left( 2 \right)+25=0$
$\Rightarrow 1+4-22-8+25=0$
$\Rightarrow \left( 1+4 \right)+\left( -22-8 \right)+25=0$
$\Rightarrow 5-30+25=0$
$\Rightarrow -25+25=0$
$\Rightarrow 0=0$
$\therefore$ LHS $=\;$ RHS
The result attained is correct.