# math

Find the centre of the circle that passes through the points p(-9,5), q(1,5), and r(-2,-2)

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1. consider the two chords PQ and PR
the centre must be the intersection of the right-bisector of these two chords.

midpoint of PQ= (-4,5)
slope of PQ = 0
equation of right-bisector of PQ is x = -4
(no real work required for this one, notice that PQ was a horizontal line, so the right-bisector must be the vertical line running through its midpoint)

midpoint of PR = (-11/2 , 3/2)
slope of PR = -7/7 = -1
so slope of perpendicular = +1
equation: x - y = C
but (-11/2 , 3/2) is on it
-11/2 -3/2 =c = -7
x - y = -7

sub in x = -4
-4 - y = -7
-y = -3
y = 3

the centre is (-4,3)

method2:

let the centre be (a,b)

the distance from (a,b) to each of those points must be the same ..., so

√((a+9)^2 + (b-5)^2) = √((a-1)^2 + (b-5)^2)
a^2 + 18a + 81 + b^2 - 10b + 25 = a^2 - 2a + 1 + b^2 -10b + 25
20a = -80
a = -4 , as before

√( (a+9)^2 + (b-5)^2) = √(a+2)^2 + (b+2)^2)
a^2 + 18a + 81 + b^2 - 10b + 25 = a^2 + 4a + 4 + b^2 + 4b + 4
14a -14b = -98
a - b = -7
but a=-4
-4 - b = -7
b = 3

then centre is (-4,3) , as before

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posted by Reiny

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