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