Find the equation of tangent to the circle x2 + y2 – 4x + 3y + 2 = 0

1.	Find the equation of tangent to the circle x2 + y2 – 4x + 3y + 2 = 0 at the point (4, -2)
Answer» Given equation of the circle is x² + y² – 4x + 3y + 2 = 0 Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get 2g = -4, 2f = 3, c = 2 g = -2, f = , c = 2 The equation of a tangent to the circle x² + y² + 2gx + 2fy + c = 0 at (x₁ , y₁) is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0 The equation of the tangent at (4, -2) is x(4) + y(-2) – 2(x + 4) + 3/2 (y-2) + 2 = 0 ⇒ 4x – 2y – 2x – 8 + 3/2 y – 3 + 2 = 0 ⇒ 2x – 1/2y - 9 = 0 ⇒ 4x – y – 18 = 0

Find the equation of tangent to the circle x2 + y2 – 4x + 3y + 2 = 0 at the point (4, -2)

Answer»

Given equation of the circle is x² + y² – 4x + 3y + 2 = 0

Comparing this equation with x² + y² + 2gx + 2fy + c = 0, we get

2g = -4, 2f = 3, c = 2

g = -2, f = , c = 2

The equation of a tangent to the circle x² + y² + 2gx + 2fy + c = 0 at (x₁ , y₁) is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0

The equation of the tangent at (4, -2) is

x(4) + y(-2) – 2(x + 4) + 3/2 (y-2) + 2 = 0

⇒ 4x – 2y – 2x – 8 + 3/2 y – 3 + 2 = 0

⇒ 2x – 1/2y - 9 = 0

⇒ 4x – y – 18 = 0

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Find the equation of tangent to the circle x2 + y2 – 4x + 3y + 2 = 0 at the point (4, -2)

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