Question

Answer the following :

Find the equations of the tangents to the circle x² + y² – 2x + 8y – 23 = 0 having slope 3

Answer 1

Let the equation of the tangent with slope 3 be y = 3x + c

i.e., 3x – y + c = 0  ...(1)

The centre of the circle x² + y² – 2x + 8y – 23 = 0 is

C (1, – 4) and its radius = `sqrt(1 + 16 + 23) = sqrt(40)`.

Since line (1) is tangent to the circle, the length of perpendicular from the centre C (1, - 4) to the line is equal to radius.

∴ `|(3(1) + (-1)(-4) + "c")/sqrt(3^2 + (-1)^2)| = sqrt(40)`

∴ `|(3 + 4 + "c")/sqrt(10)| = sqrt(40)`

∴ 7 + c = ± 20

∴ 7 + c = 20 or 7 + c = – 20

∴ c = 13 or c = – 27

∴ from (1), the equations of the tangents are

3x – y + 13 = 0 and 3x – y – 27 = 0.