Math CentralQuandaries & Queries


Question from Renson, a student:

A circle passes through (4,2) and (11,-5) with center on the line 3y-5x+41=0. Determine its equation in standard form.


Suppose the center is (h,k), then since the center is on the line

3k - 5h + 41 = 0


k = 5/3 h - 41/3            (equation 1).

Since (4,2) and (11,-5) are on the circle the distance from (4,2) to (h,k) is the same as the distance from (11,-5) to (h,k). Use the distance formula to write these two distances and set them equal. Substitute k = 5/3 h - 41/3, expand and simplify the resulting equation (a quadratic in h) and solve for h. Equation 1 then gives you k and the distance from (h,k) to either of the given points is the radius r.


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