Perpendicular bisectors and centres of circles

**stewpot** · March 6th 2009, 06:09 AM

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

**Soroban** · March 6th 2009, 06:49 AM

Hello, stewpot!

Your equations for the perpendicular bisectors are off . . .

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

Find the equation of the circle.

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

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

Now try it . . .

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