Question

Show that the line 7x − 3y − 1 = 0 touches the circle x² + y² + 5x − 7y + 4 = 0 at point (1, 2)

Answer 1

Given equation of the circle is

x² + y² + 5x − 7y + 4 = 0

Comparing this equation with

x² + y² + 2gx + 2fy + c = 0, we get

2g = 5, 2f = – 7, c = 4

∴ g = `5/2, "f" = -7/2, "c" = 4`

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 (1, 2) is

`x(1) + y(2) + 5/2(x + 1) - 7/2(y + 2) + 4` = 0

∴ `x + 2y + 5/2x + 5/2 - 7/2y - 7 + 4` = 0

∴ `7/2x - 3/2y - 1/2` = 0

∴ 7x – 3y – 1 = 0, which is same as the given line

∴ The line 7x – 3y – 1 = 0 touches the given circle at (1, 2).