Question

Find the equation of the pair of tangents drawn to the circle x² + y² – 2x + 4y = 0 from the point (0, 1)

Answer 1

Given circle is S = x² + y² – 2x + 4y = 0 .......(i) 

Let P ≡ (0, 1) 

For point P, S₁ = 0² + 1² – 2.0 + 4.1 = 5 

Clearly P lies outside the circle 

and T ≡ x . 0 + y . 1 – (x + 0) + 2 (y + 1) 

i.e. T ≡ –x + 3y + 2. 

Now equation of pair of tangents from P(0, 1) to circle (1) is SS₁ = T₂ 

or 5 (x² + y² – 2x + 4y) = (– x + 3y + 2)² 

or 5x² + 5y² – 10x + 20y = x² + 9y² + 4 – 6xy – 4x + 12y 

or 4x² – 4y² – 6x + 8y + 6xy – 4 = 0 

or 2x² – 2y² + 3xy – 3x + 4y – 2 = 0 .......(ii) 

Note : Separate equation of pair of tangents : From (ii), 2x² + 3(y – 1) x – 2(2y² – 4y + 2) = 0

∴ Separate equations of tangents are x – 2y + 2 = 0 and 2x + y – 1 = 0

Comments

In this process,
T is given by,
T ≡ xx1 + yy1 + f(x+x1) + g(y+y1) + C
Where equation of circle is,
x^2 + y^2 +2fx + 2gy + c = 0.
Here,
x1 = 0, y1 = 1 & f = -1, g = 2, c = 0.