Maths question about equations of circles

Question

A circle which touches the straight lines L:7x-4y-52=0 at the point A (8,1) and the circle passes through the point B(-7,4)

find the centre of the circle

Answer 1

7x-4y-52= 0 is a line tangent to a circle

4y = 7x-52

y = 7x/4- 13, 7/4 = the tangent line's slope

the circle includes a point (8,1), but it's also on the tangent line

as 7(8)-4(1)-52 = 56-4-52 = 0

it may help to graph the line and point

(-7,4) ia another point on the circle

general formula for a circle is

(x-h)^2 + (y-k)^2 = r^2 where (h,k) = the circle's center and r= the radius

(-7-h)^2 + (4-k)^2 = r^2

(8-h)^2 + (1-k)^2 = r^2

49+14h +h^2 + 16 -8k + k^2 = r^2

h^2 +14h + k^2 -8k = r^2 -65

h^2 - 16h + k^2 -2k = r^2 -65

subtract

30h -6k = 0

5h = k

y = 5x

perpendicular line through the center is y=-4x/7 -37/7 and through (8,1)

5x = -4x/7 -39/7

35x = -4x -39

39x =39

x = 1

y =5x = 5

(1,5) is the center of the circle

from (1,5) to (8,1) is 8 = radius