What is the equation of the tangent line that touches the circle x squared plus 10x plus y squared -2y -39 equals 0 at the point of 3 2 on the Cartesian plane showing work?

First find the slope of the circle's radius as follows:-

Equation of circle: x^2 +10x +y^2 -2y -39 = 0

Completing the squares: (x+5)^2 + (y-1)^2 -25 -1 -39 = 0

So: (x+5)^2 +(y-1)^2 = 65

Centre of circle: (-5, 1) and point of contact (3, 2)

Slope of radius: (1-2)/(-5-3) = 1/8 which is perpendicular to the tangent line

Slope of tangent line: -8

Tangent equation: y-2 = -8(x-3) => y = -8x+26

Tangent equation in its general form: 8x+y-26 = 0