Question

How do you find the centre of a circle given two points $?$

Answer 1

Step 1. Let $(a,b)$ is the centre and $r$ is the radius of the circle.

If $(x_1,y_1)$ and $(x_2,y_2)$ are two points, then we use the below equations to find the centre of the circle.

${\mathbf{(}{\mathit{x}}_{\mathbf{1}}\mathbf{}\mathbf{–}\mathbf{}\mathit{a}\mathbf{)}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{(}{\mathit{y}}_{\mathbf{1}}\mathbf{}\mathbf{–}\mathbf{}\mathit{b}\mathbf{)}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{r}}^{\mathbf{2}}$ and ${\mathbf{(}{\mathit{x}}_{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}\mathit{a}\mathbf{)}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{(}{\mathit{y}}_{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}\mathit{b}\mathbf{)}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{r}}^{\mathbf{2}}$.

For Example; Let $\left(5,4\right)$ and $\left(3,6\right)$ be two points and let the radius be $10$, then

If $x=5,y=4$ then ${\left(5-a\right)}^2+{\left(4-b\right)}^2=100$ …(1)

If $x=3,y=6$ then ${\left(3-a\right)}^2+{\left(6-b\right)}^2=100$ …(2)

Expanding equation (1) we get

$25-10a+a^2+16-8b+b^2=100$ …(3)

Expanding equation (2), we get

$9-6a+a^2+36-12b+b^2=100$ …(4)

Step 2. Subtracting equation (4) from equation (3)

$16-4a-20+4b=0$

$\Rightarrow$ $-4a+4b=4$

$\Rightarrow$ $b=a+1$

Step 3. Substitute the value of $b$ in equation (3)

$25-10a+a^2+16-8\left(a+1\right)+\hspace{0.17em}{\left(a+1\right)}^2=100$

$\Rightarrow$$25-10a+\hspace{0.17em}{a}^2+16-8a+a^2+\hspace{0.17em}2a+1=100$

$\Rightarrow$ $2a^2-16a-66=0$

$\mathbf{\therefore }{\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{8}\mathit{a}\mathbf{}\mathbf{-}\mathbf{}\mathbf{33}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$ is the required equation of circle.

Hence, we can find the Centre of a circle by solving above two equations.