Question

Answer the following :

If the tangent at (3, –4) to the circle x² + y² = 25 touches the circle x² + y² + 8x – 4y + c = 0, find c

Answer 1

The equation of a tangent to the circle

x² + y² = r² at (x₁, y₁) is xx₁ + yy₁ = r²

Equation of the tangent at (3, – 4) is

x(3) + y(–4) = 25

∴ 3x – 4y – 25 = 0 …(i)

Given equation of circle is

x² + y² + 8x – 4y + c = 0

Comparing this equation with

x² + y² + 2gx + 2fy + c = 0, we get

2g = 8, 2f = – 4

∴ g = 4, f = – 2

∴ C ≡ (– 4, 2) and r = `sqrt(4^2 + (-2)^2 - "c") = sqrt(20 - "c")`

Since line (i) is a tangent to this circle also, the perpendicular distance from C(– 4, 2) to line (i) is equal to radius r.

∴ `|(3(-4) + (-4)(2) - 25)/sqrt(3^2 + 4^2)| = sqrt(20 - "c")`

∴ `|(-45)/sqrt(25)| = sqrt(20 - "c")`

∴ `|(-45)/5| = sqrt(20 - "c")`

∴ `|–9| = sqrt(20 - "c")`

∴ 81 = 20 – c   ...[Squaring both the sides]

∴ c = –61