JEE Main · 2015 · 04 Apr 2015 · Offline · Circle

The number of common tangents to the circles

x^{2} + y^{2} - 4 x - 6 y - 12 = 0

and

x^{2} + y^{2} + 6 x + 18 y + 26 = 0

, is :

(A)2
(B)3✓ Correct
(C)4
(D)1

Answer: (B)

Solution

Read each circle off its equation. The first,

x^{2} + y^{2} - 4 x - 6 y - 12 = 0

, has centre

(2, 3)

and radius

4 + 9 + 12 = 5

; the second,

x^{2} + y^{2} + 6 x + 18 y + 26 = 0

, has centre

(- 3, - 9)

and radius

9 + 81 - 26 = 8

. The centres are

5^{2} + 1 2^{2} = 13

apart, exactly

5 + 8

, so the circles touch externally. Two circles touching externally share three common tangents -- two external ones and the one at the point of contact.

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

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