Question

If the circles x² + y² + 2x + 2ky + 6 = 0 and x² + y² + 2ky + k = 0 intersect orthogonaly then k is

(a) 2 or –3/2

(b) –2or–3/2

(c) 2 or 3/2

(d) – 2or3/2

Answer 1

Correct option is (a) 2 or –3/2

Condition for two circles to intersect at right angles is 2g₁g₂ + 2f₁f₂ = c₁ + c₂

Here two circles are x² + y² + 2x + 2ky + 6 = 0 and x² + y² + 2ky + k = 0