The number of common tangents, to the circles $x^{2}+y^{2}-18 x-15 y+131=0$ and $\mathrm{x}^{2}+\mathrm{y}^{2}-6 \mathrm{x}-6 \mathrm{y}-7=0$, is:

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Circle

The number of common tangents, to the circles

x^{2} + y^{2} - 18 x - 15 y + 131 = 0

and

x^{2} + y^{2} - 6 x - 6 y - 7 = 0

, is:

(A)3✓ Correct
(B)2
(C)1
(D)4

Answer: (A)

Solution

Completing the square, the first circle is

(x - 9)^{2} + (y - \frac{15}{2})^{2} = \frac{25}{4}

, centre

C_{1} = (9, \frac{15}{2})

, radius

r_{1} = \frac{5}{2}

, and the second is

(x - 3)^{2} + (y - 3)^{2} = 25

, centre

C_{2} = (3, 3)

, radius

r_{2} = 5

. The distance between centres is

C_{1} C_{2} = 6^{2} + (\frac{9}{2})^{2} = 36 + \frac{81}{4} = \frac{15}{2}

. Since

C_{1} C_{2} = r_{1} + r_{2}

, the circles touch externally, so they have exactly three common tangents.

JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Single correct option

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