Question

Answer the following :

Find the equation of tangent to Circle x² + y² – 6x – 4y = 0, at the point (6, 4) on it

Answer 1

Given equation of the circle is

x² + y² – 6x – 4y = 0

Comparing this equation with

x² + y² + 2gx + 2fy + c = 0, we get

2g = –6, 2f = –4, c = 0

∴ g = –3, f = –2, c = 0

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 (6, 4) is

x(6) + y(4) – 3(x + 6) – 2(y + 4) + 0 = 0

∴ 6x + 4y – 3x – 18 – 2y – 8 = 0

∴ 3x + 2y – 26 = 0

Answer 2

Given equation of the circle is

x² + y² – 6x – 4y = 0

∴ x(x – 6) + y(y – 4) = 0, which is in diameter form where (0, 0) and (6, 4) are endpoints of diameter.

Slope of OP = `(4 - 0)/(6 - 0) = 2/3`

Since, OP is perpendicular to the required tangent.

∴ Slope of the required tangent =  `(-3)/2`

∴ the equation of the tangent at (6, 4) is

y – 4 = `(-3)/2(x - 6)`

∴ 2(y – 4) = –3(x – 6)

∴ 2y – 8 = –3x + 18

∴ 3x + 2y – 26 = 0