Equation of the tangent line to the circle?? help?

Question

find an equation of the tangent line to the circle whose equation os x^2 - 4x +y^2 +2y = 0 at a point in the fourth quadrant where x=4

i need solutions thanks :))

Answer 1

2x – 4 + 2y(dy/dx) + 2(dy/dx) = 0

dy/dx (2y + 2) = 4 – 2x

dy/dx = (2y + 2)/(4 – 2x)

4² - 4(4) + y² + 2y = y(y + 2) = 0

y = -2

y = mx + b; (4, -2)

m = (2(-2) + 2)/(4 – 2(4)) = -2/-4 = ½

-2 = (1/2)(4) + b

-2 = 2 + b

-4 = b

equation: y = (1/2)x - 4

Answer 2

If x = 4 then,

16 -16 +y^2 +2y = 0 so y=0 or -1 but for the point to be in Quad 4, y = -1. So the pint of contact is (4, -1)

Center of the circle is (2, -1) and so the slope of the radius at that point is 0. Hence radius is // x_axis and tangent at tha point must be // y-axis so its eqn is x = 4 itself.