A circle passes through the center of a (- 2,1) B (5,0) on the straight line x-3y-10 = 0. The equation for finding a circle is expressed by the slope A circle passes through the center of a (- 2,1) B (5,0) and is on the straight line x-3y-10 = 0. The equation of circle is solved by the method of slope

The slope of AB line is: (0-1) / (5 + 2) = - 1 / 7, take the midpoint of AB as D (3 / 2,1 / 2), make the vertical line of AB through D, and the slope is 7,
Then: Y-1 / 2 = 7 (x-3 / 2)
The result is: 7x-y-10 = 0
Then it is combined with x-3y-10 = 0
Get the center of the circle (1, - 3)
Find the distance to point a, d = 5
So the equation is: (x-1) square + (y + 3) square = 25

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