# Perpendicular bisectors and centres of circles

• Mar 6th 2009, 06:09 AM
stewpot
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
• Mar 6th 2009, 06:49 AM
Soroban
Hello, stewpot!

Your equations for the perpendicular bisectors are off . . .

Quote:

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 . . .