# Thread: Perpendicular bisectors and centres of circles

1. ## Perpendicular bisectors and centres of circles

Hi hope someone can help me?

The question concerns three points A (3,1) B (-1,-11) and C (-5,9) that a circle passes through. I've managed to find the slope, midpoint and perpendicular bisector of AB (y=-x-14/3) and the same for AC(y=x+4). However when I try to find the centre and radius, I can't get the circles equation to balance. (The coordinates of each of the points A,B and C should balance to satisfy the equation).

Would be very happy if someone could point me in the right direction.

Many thanks

Stewpot

2. Hello, stewpot!

Your equations for the perpendicular bisectors are off . . .

Three points on a circle: .$\displaystyle A (3,1),\;B (\text{-}1,\text{-}11),\;C (\text{-}5,9)$

Find the equation of the circle.

The midpoint of $\displaystyle AB$ is: .$\displaystyle (1,\text{-}5)$
The slope of $\displaystyle AB$ is: .$\displaystyle m_{AB}\,=\,3$
The perpendicular bisector of $\displaystyle AB$ has point$\displaystyle (1,\text{-}5)$ and slope $\displaystyle \text{-}\tfrac{1}{3}$
. . ts equation is: .$\displaystyle y +5 \:=\:\text{-}\tfrac{1}{3}(x - 1) \quad\Rightarrow\quad \boxed{y \:=\:\text{-}\tfrac{1}{3}x - \tfrac{14}{3}}$

The midpoint of $\displaystyle AC$ is: .$\displaystyle (\text{-}1,5)$
The slope of $\displaystyle AC$ is: .$\displaystyle m_{AC} \,=\,\text{-}1$
The pependicular bisector of $\displaystyle AC$ has point $\displaystyle (\text{-}1,5)$ and slope $\displaystyle +1$
. . Its equation is: .$\displaystyle y - 5 \:=\:1(x+1) \quad\Rightarrow\quad\boxed{ y \:=\:x+6}$

Now try it . . .