# Equation of chord AB of circle $x^2 + y^2 = 2$  passing through P(2 , 2) such that PB/PA = 3…

Problem: Equation of chord AB of circle $x^2 + y^2 = 2$  passing through P(2 , 2) such that PB/PA = 3, is given by

(A) x = 3 y

(B) x = y

(C) y – 2 = √3(x – 2)

(D) none of these

Sol. Any line passing through (2, 2) will be of the form $\large \frac{y-2}{sin\theta} = \frac{x-2}{cos\theta} = r$

When this line cuts the circle x2 + y2 =2 ,

(rcosθ + 2)2 + (r sinθ + 2)2 = 2

⇒  r2 + 4(sinθ+ cosθ)r + 6 = 0

$\large \frac{PB}{PA} = \frac{r_2}{r_1}$ , now if r1 = α , r2 = 3α ,

then 4α = – 4(sinθ + cosθ), 3α2 = 6

⇒ sin2θ = 1

⇒ θ = π/4

So required chord will be y – 2 = 1 ( x –2)

y = x.