Determine the equation of the circle passes through the poin

Determine the equation of the circle passes through the point (-4,1) and its center is the midpoint on the line segment joining the centers of two circles x²+y²-6x-4y+4=0 and  x²+y²-14x+47=0.

I honestly don\'t know how to even go about this problem.. detailed explanations would be greatly appreciated! Thank you so much!

Solution

General equation of circle is x² + y² + 2hx + 2ky + c = 0

==> Centre = (-h , -k)

Given equations of circles are x²+y²-6x-4y+4=0 ==> Centre = (6/2 , 4/2) = (3 , 2)

and x²+y²-14x+47=0 ==> Centre = (14/2 , 0) = (7 , 0)

Mid point of (3 , 2) and (7 , 0) is ( (3 +7)/2 , (2 + 0)/2) = (10/2 , 2/2) = (5 , 1)

Distance between (5 , 1) and given point (-4 , 1) is nothing but radius of circle

==> r = [(5 -(-4))² + (1 - 1)²] = 9² = 9

Equation of circle with radius 9 and centre (5 , 1) is

(x -5)² + (y -1)² = 9²

==> x² -10x + 25 + y² -2y +1 = 81

==> x² + y² -10x -2y -55 = 0 is required equation of circle.