1.

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

Answer»

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

xx1 + yy1 – 4\(\frac{(x+x_1)}{2}\) + 4\(\frac{(x+y_1)}{2}\) – 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



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