Question

Find the equations of the two tangents from the point (5, -3) to the circle x² + y² - 4x + 8y + 12 = 0.

Answer 1

Formula:

Let the equation of a circle be

S ≡ x² + y² + 2gx + 2fy + c = 0

Then the equation of the pair of tangents to the circle S = 0 from a point (h, k) is given by

SS₁ = T² ,

where S = x² + y² + 2gx + 2fy + c

S₁ = h² + k² + 2gh + 2fk + c

T = xh + yk + g(x + h) + f(y + k) + c

From the above relation, an equation of pair of tangents can be found, which in further can calculations can be represented as the product of two tangents.

{ Refer to the diagram added }

Solution:

Since the tangents to the circle S = 0 are drawn from the point (5, - 3), the pair of tangents can be found from the above relation as follows,

(x² + y² - 4x + 8y + 12) (5² + 3² - 20 - 24 + 12) = {5x - 3y - 2(x + 5) + 4(y - 3) + 12}²

or, 2 (x² + y² - 4x + 8y + 12) = (3x + y - 10)²

or, 2x² + 2y² - 8x + 16y + 24 = 9x² + y² + 100 + 6xy - 60x - 20y

or, 7x² + 6xy - y² - 52x - 36y + 76 = 0

or, 7x² + (6y - 52)x + (- y² - 36y + 76) = 0

∴ x = [- (6y - 52) ± √{(6y - 52)² - 4 * 7 * (- y² - 36y + 76)}] / (2 * 7)

= {- 6y + 52 ± √(36y² - 624y + 2704 + 28y² + 1008y - 2128)} / 14

= {- 6y + 52 ± √(64y² + 384y + 576)} / 14

= [- 6y + 52 ± √{(8y + 24)²}] / 14

or, 14x = - 6y + 52 ± (8y + 24)

or, 7x = - 3y + 26 ± (4y + 12)

This gives two equations,

7x = - 3y + 26 + 4y + 12

or, 7x - y = 38

and 7x = - 3y + 26 - 4y - 12

or, 7x + 7y = 14

or, x + y = 7

Therefore, the required tangents are

7x - y = 38 , x + y = 2