Question

Find the equation of the tangent to the circle x² + y² – 4x + 4y – 8 = 0 at (-2, -2).

Answer 1

The equation of the tangent to the circle x² + y² – 4x + 4y – 8 = 0 at (x₁, y₁) is

xx₁ + yy₁ – 4`((x + x_1))/2 + 4((y + y_1))/2 - 8 = 0`

Here (x₁, y₁) = (-2, -2)

⇒ x(-2) + y(-2) – 2(x – 2) + 2(y – 2) – 8 = 0

⇒ -2x – 2y – 2x + 4 + 2y – 4 – 8 = 0

⇒ -4x – 8 = 0

⇒ x + 2 = 0