The circle should pass through the three points (-7,4),(-4,5) and (0,3)

Call the center of the circle P. Then PA=PB=PC since all radii of a circle are congruent.

PA=PB implies that P lies on the perpendicular bisector of AB; similarly P lies on the perpendicular bisector of BC and AC.

The perpendicular bisector of a segment goes through the midpoint of the segment, and the slope is the opposite reciprocal of the slope of the segment.

(i) The `_|_` bisector of AB: The midpoint of AB is `((-7+-4)/2,(4+5)/2)=(-11/2,9/2)` . The slope of AB is `1/3` , so the slope of the `_|_` bisector is -3. Then the equation for the `_|_` bisector of AB is `y-9/2=-3(x+11/2)==> y=-3x-12`

(ii) The `_|_` bisector of BC: The midpoint of BC is `((-4+0)/2,(5+3)/2)=(-2,4)` The slope of BC is `-1/2` , so the slope of the `_|_` bisector is 2.

Then the equation of the `_|_` bisector is `y-4=2(x+2)==>y=2x+8`

(iii) The perpendicular bisectors meet when `-3x-12=2x+8==>x=-4` and `y=0`

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**The center of the circle going through the given points is (-4,0)**

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** You could check by finding the `_|_` bisector of AC: the midpoint is (-3.5,3.5) and the slope of the `_|_` bisector will be 7 giving you `y-3.5=7(x+3.5)==> y=7x+28`

Let `-3x-12=7x+28==>x=-4,y=0` as before.

This is the circumcircle of triangle ABC, and P is the cicumcenter.

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