Find the equation of the tangent and normal to the circle x2 + y2 –

1.	Find the equation of the tangent and normal to the circle x2 + y2 – 6x + 6y – 8 = 0 at (2, 2).
Answer» The equation of the tangent to the circle x² + y² + 2 gx + 2fy + c = 0 at (x₁, y₁) is xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 9 So the equation of the tangent to the circle x² + y² – 6x + 6y – 8 = 0 at (x₁, y₁) is xx₁ + yy₁ - (6(x + x₁)/2) + (6(y + y₁)/2) - 8 = 0 (i.e) xx₁ + yy₁ – 3(x + x₁) + 3(y + y₁) – 8 = 0 Here (x₁, y₁) = (2, 2) So equation of the tangent is x(2) + y(2) – 3(x + 2) + 3(y + 2) – 8 = 0 (.i.e) 2x + 2y – 3x – 6 + 3y + 6 – 8 = 0 (i.e) -x + 5y – 8 = 0 or x – 5y + 8 = 0 Normal is a line ⊥r to the tangent So equation of normal circle be of the form 5x + y + k = 0 The normal is drawn at (2, 2) ⇒ 10 + 2 + k = 0 ⇒ k = -12 So equation of normal is 5x + y – 12 = 0

Find the equation of the tangent and normal to the circle x2 + y2 – 6x + 6y – 8 = 0 at (2, 2).

Answer»

The equation of the tangent to the circle x² + y² + 2 gx + 2fy + c = 0 at (x₁, y₁) is

xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 9

So the equation of the tangent to the circle

x² + y² – 6x + 6y – 8 = 0 at (x₁, y₁) is

xx₁ + yy₁ - (6(x + x₁)/2) + (6(y + y₁)/2) - 8 = 0

(i.e) xx₁ + yy₁ – 3(x + x₁) + 3(y + y₁) – 8 = 0

Here (x₁, y₁) = (2, 2)

So equation of the tangent is

x(2) + y(2) – 3(x + 2) + 3(y + 2) – 8 = 0

(.i.e) 2x + 2y – 3x – 6 + 3y + 6 – 8 = 0

(i.e) -x + 5y – 8 = 0 or x – 5y + 8 = 0

Normal is a line ⊥r to the tangent

So equation of normal circle be of the form 5x + y + k = 0

The normal is drawn at (2, 2)

⇒ 10 + 2 + k = 0 ⇒ k = -12

So equation of normal is 5x + y – 12 = 0