What is the tangent line equation of the circle x2 plus y2 -6x plus 4y -7 equals 0 passing through the point 1 2?

Differentiate the circle equation to find the slope at any given point.

x^2 + y^2 -6x + 4y -7 = 0

2x + 2y(dy/dx) - 6 + 4dy/dx =0

2x - 6 + (2y + 4) dy/dx =0

dy/dx = (6 - 2x) / (2y + +4)

At the point ( 1,2)

dy/dx = (6 - 2(1)) / (2(2) + 4)

dy/dx = 3 / 8 The slope /gradient of the tangent line.

At the point ( 1,2)

y - 2 = (3/8)(x - 1)

y - 2 = 3x/8 - 3/8

y = 3x/8 + 13/8

or

8y = 3x + 13

or

8y - 3x = 13

or

8y - 3x - 13 = 0

8