Question 1015160
(x+2k)^2 + (y-3k)^2 = 25 pass through the point (1,0)
find value of k
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(-2k,3k) is the center of the given circle, call it C.
The distance from (-2k,3k) to (1,0) = 5
--> 25 = (-2k-1)^2 + (3k)^2 = 4k^2 + 4k + 1 + 9k^2
13k^2 + 4k - 24 = 0
*[invoke solve_quadratic_equation 13,4,-24]
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k = the 2 values above.
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There are 2 circles of r = 5 that fit.