# How do you write the general form given a circle that passes through the given points. P(-2, 0), Q(1, -3), R(-2, -6)?

Apr 15, 2016

General of equation of the circle would be ${x}^{2} + {y}^{2} + 4 x + 6 y + 4 = 0$

#### Explanation:

We are given three points $P \left(- 2 , 0\right)$, $Q \left(1 , - 3\right)$ and $R \left(- 2 , - 6\right)$.

The perpendicular bisectors of any two lines joining these points will intersect at the center of the circle passing through these points.

Slope of line joining PQ is $\frac{- 3 - 0}{1 - \left(- 2\right)} = - \frac{3}{3} = - 1$ and mid point of PQ is $\left(\frac{- 2 + 1}{2} , \frac{0 - 3}{2}\right)$ or $\left(- \frac{1}{2} , - \frac{3}{2}\right)$. Hence its perpendicular bisector will pass through $\left(- \frac{1}{2} , - \frac{3}{2}\right)$ and have a slope of $- \frac{1}{-} 1 = 1$.

Hence equation of perpendicular bisector of PQ is $y - \left(- \frac{3}{2}\right) = 1 \times \left(x - \left(- \frac{1}{2}\right)\right)$ or $y + \frac{3}{2} = x + \frac{1}{2}$ or $x = y + 1$.

Slope of line joining QR is $\frac{- 6 - \left(- 3\right)}{- 2 - 1} = - \frac{3}{-} 3 = 1$ and mid point of QR is $\left(\frac{- 2 + 1}{2} , \frac{- 6 - 3}{2}\right)$ or $\left(- \frac{1}{2} , - \frac{9}{2}\right)$. Hence its perpendicular bisector will pass through $\left(- \frac{1}{2} , - \frac{9}{2}\right)$ and have a slope of $- \frac{1}{1} = - 1$.

Hence equation of perpendicular bisector of PQ is $y - \left(- \frac{9}{2}\right) = - 1 \times \left(x - \left(- \frac{1}{2}\right)\right)$ or $y + \frac{9}{2} = - x - \frac{1}{2}$ or $x + y + 5 = 0$.

Solving $x = y + 1$ and $x + y + 5 = 0$ gives us $\left(- 2 , - 3\right)$, which is center $O$ and radius will be its distance with say $P \left(- 2 , 0\right)$, which is $3$

Hence equation of circle will be ${\left(x - \left(- 2\right)\right)}^{2} + {\left(y - \left(- 3\right)\right)}^{2} = {3}^{2}$

or ${\left(x + 2\right)}^{2} + {\left(y + 3\right)}^{2} = 9$ or ${x}^{2} + 4 x + 4 + {y}^{2} + 6 y + 9 = 9$ and simplifying it general for of equation of circle would be ${x}^{2} + {y}^{2} + 4 x + 6 y + 4 = 0$