Find the standard equation of the circle whose center is on the straight line 2x-y-3 = 0 and passes through points (5,2) and (3, - 2)
The center of the circle is on the straight line 2x-y-3 = 0, y = 2x-3
Let the center of the circle be (a, 2a-3)
(5-a)^2+(2-2a+3)^2=r^2
(3 - a)^2 +(-2-2a+3)^2=r^2
The solution is a = 2
r^2=10
Center of circle (2,1)
Equation (X-2) ^ 2 + (Y-1) ^ 2 = 10
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