# How do you find the equation of the circle passing through the points P(x_1,y_1), Q(x_2,y_2), R(x_3,y_3)?

## How do you find the equation of the circle passing through the points $P \left({x}_{1} , {y}_{1}\right) , Q \left({x}_{2} , {y}_{2}\right) , R \left({x}_{3} , {y}_{3}\right)$?

##### 1 Answer
Feb 7, 2018

Please see below.

#### Explanation:

There are two ways of doing this.

First Method $-$ Find the equation of perpendicular bisector of segment joining one pair of points, say . Then find the equation of perpendicular bisector of segment joining another pair of points, say $\left({x}_{2} , {y}_{2}\right)$ and $\left({x}_{3} , {y}_{3}\right)$.

Find the point of intersection of two perpendicular bisectors by solving their simultaneous equations to get the center and then its distance to any of the three points is radius.

If center is $\left(h , k\right)$ and radius is $r$, equation of circle is ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$.

Second Method $-$ Assume the equation of circle to be ${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$. As we put the values of abscissa and ordinates from coordinates of three points $\left({x}_{1} , {y}_{1}\right)$, $\left({x}_{2} , {y}_{2}\right)$ and $\left({x}_{3} , {y}_{3}\right)$, we get three simultaneous equations in $g , f$ and $c$.

Solving them gives values of $g , f$ and $c$ and putting them in ${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$, we get equation of circle.