Question

Let the circles S ≡ x² + y² - 12 = 0 and S' ≡  x² + y² - 5x + 3y - 2 = 0 intersect at points P and Q  Tangents are drawn to the circle S ≡ x² + y² - 12 = 0 at points P and Q. The point of intersection of these tangents is

(a)   (6, - 18/5)

(b)  (6, 18/5)

(c)  (-6, 18/5)

(d)  (-6, - 18/5)

Answer 1

Correct option (a)   (6, - 18/5)

Explanation :

 The common chord of the circles is L ≡  S -S' ≡ 5x - 3y - 10 = 0. Note that PQ is L = 0. Suppose the tangents at points P and Q meet in T(h, k). Therefore, the equation of the chord PQ is

hx + ky = 12  ...(1)

However,

 L ≡ 5x - 3y - 10 = 0  ...(2)

 is PQ. Therefore, from Eqs. (1) and (2), we get h/5 k/ 3 12/10 = 6/5. Hence

h = 6,k = -18/5

Therefore

T(h,k) = (6, -18/5)