# The perpendicular bisector of a chord of a circle passes through its centre. Find the centre passing through the points P(5,7), Q(7,1) and R(-1,5) by finding the perpendicular bisectors of PQ and QR and solving them simultaneously We are asked to find the center of a circle that passes through the points P(5,7),Q(7,1) and R(-1,5):

(1) First we find the equation of the perpendicular bisector of PQ:

To find the equation of a line we need a point and the slope.

(a) The point we require is...

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We are asked to find the center of a circle that passes through the points P(5,7),Q(7,1) and R(-1,5):

(1) First we find the equation of the perpendicular bisector of PQ:

To find the equation of a line we need a point and the slope.

(a) The point we require is the midpoint of `bar(PQ)` . The coordinates of the midpoint of a segment are the means of the coordinates of the endpoints; if the endpoints are (a,b) and (c,d) then the midpoint is `((a+c)/2,(b+d)/2)` . Here the midpoint is (6,4).

(b) The slope of the line through P and Q is `m_(bar(PQ))=(7-1)/(5-7)=-3` . The perpendicular bisector is perpendicular to this line so its slope is the opposite reciprocal. The slope of the perp. bisector is `m=1/3`

(c) We have a point (6,4) and a slope `m=1/3` . The equation of the line is `y-4=1/3(x-6)==>y=1/3x+2`

(2) We also need the equation of the perpendicular bisector through `bar(QR)` . We proceed as in (1):

(a) The midpoint is (3,3)

(b) The slope through Q and R is `m_(bar(QR))=(5-1)/(-1-7)=-1/2`

The slope of the perpendicular bisector is 2.

(c) We have the point (3,3) and the slope m=2. The equation of the perpendicular bisector is y-3=2(x-3) ==> y=2x-3

(3) We now solve the system `y=1/3x+2,y=2x-3` to find the center. We can use substitution:

`1/3x+2=2x-3`

`x+6=6x-9`

`5x=15`

x=3 ==> y=3

The center of the circle is (3,3).

The graph:

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