Find the coordinates of the circumcentre of the triangle whose vertices are (2,-2),(8,6) and (8,-2). Also find the circum radius of the triangle.
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The circumcentre of the triangle is the centre of the circle, O,  passing through all three points.

Standard equation for circle: (x-h)^2+(y-k)^2=r^2. (h,k) is O.

(2-h)^2+(2+k)^2=r^2=(8-h)^2+(6-k)^2=(8-h)^2+(2+k)^2.

So, (6-k)^2-(2+k)^2=0; (6-k-2-k)(6-k+2+k)=0=8(4-2k) and k=2.

(2-h)^2-(8-h)^2=0=-6(10-2h), and h=5, so O is (5,2).

r^2=(2-5)^2+(2+2)^2=9+16=25, and r=5, the circumradius.

Equation of circle is (x-5)^2+(y-2)^2=25.

CHECK

Put x=2 or 8: 9+(y-2)^2=25; y-2=±4, so y=-2, 6 are solutions and (2,-2), (8,-2) and (8,6) are points on the circumference.

by Top Rated User (761k points)

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