Question

If P and Q are the points of intersection of the circles x² + y² + 3x + 7y + 2p – 5 = 0 and x² + y² + 2x + 2y – p² = 0, then there is a circle passing through P, Q and (1, 1) for
(a) all except one value of p
(b) all except two values of p
(c) exactly one value of p
(d) all values of p

Answer 1

(a) : The radical axis, which in the case of intersection of the circles is the common chord, of the circles
S₁ : x² + y² + 3x + 7y + 2p – 5 = 0 and
S₂ : x² + y² + 2x + 2y – p² = 0 is S₁ – S₂ = 0
⇒ x + 5y + 2p – 5 + p² = 0 ...(i)
If there is a circle passing through P, Q and (1, 1)
it’s necessary and sufficient that (1, 1) doesn’t lie on
PQ, i.e., 1 + 5 + 2p – 5 + p² ≠ 0
⇒ p² + 2p + 1 ≠ 0 ⇒ (p + 1)² ≠ 0
∴ p ≠ –1
Thus for all values of p except ‘–1’ there is a circle passing through P, Q and (1, 1).