If the circles x2 + y2 + 2x + 2ky + 6 = 0 and x2 + y2 + 2ky + k =

1.	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) –2 or –3/2(c) 2 or 3/2(d) – 2 or 3/2
Answer» Correct option (a) 2 or –3/2 Explanation: Condition for two circles to intersect at right angles is 2g₁ g₂ + 2f₁ f₂ = c₁ + c₂ Here two circles are x²+ y²+ 2x + 2ky + 6 = 0 and x²+ y²+ 2ky + k = 0 g₁ = 1, f₁ = k c₁ = 6 g₂ = 0 f₂ = k c₂ = k 0 + 2k² = 6 + k 2k²– k – 6 = 0 2k²– 4k + 3k – 6 = 0 (2k + 3) (k – 2) = 0 k = –3/2 or k = 2

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) –2 or –3/2(c) 2 or 3/2(d) – 2 or 3/2

Answer»

Correct option (a) 2 or –3/2

Explanation:

Condition for two circles to intersect at right angles is 2g₁ g₂ + 2f₁ f₂ = c₁ + c₂

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

and x²+ y²+ 2ky + k = 0

g₁ = 1, f₁ = k c₁ = 6

g₂ = 0 f₂ = k c₂ = k

0 + 2k² = 6 + k

2k²– k – 6 = 0

2k²– 4k + 3k – 6 = 0

(2k + 3) (k – 2) = 0

k = –3/2 or k = 2

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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) –2 or –3/2(c) 2 or 3/2(d) – 2 or 3/2

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