Question

If two circle (x – 1)² + (y – 3)² = a² and x² + y² – 8x + 2y + 8 = 0 intersect in two distinct points, then

(a) 2 < a < 8

(b) a > 2

(c) a < 2

(d) a = 2

Answer 1

The correct option (a) 2 < a < 8

Explanation:

Let d is distance between centre of two circles of radii r₁ & r₂.

Then they intersect in two distinct points if |r₁ – r₂| < d < |r₁ + r₂|

Here radii of circle are a and √(16 + 1 – 8) = 3.

Distance between centre is d[(1, 3) and (+ 4, – 1)]

i.e. distance is √[(1 – 4)2 + (3 + 1)²] = √(9 + 16) = 5

∴        |a – 3| < 5 < |a + 3|

∴   – 2 < a < 8  and  a > 2

∴     2 < a < 8.