JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

Choose the correct answer:

Explanation

Circle 1 ( $C_1$ ): Center $O_1 = (9, 7.5)$ , Radius $r_1 = \sqrt{9^2 + 7.5^2 - 131} = \sqrt{81 + 56.25 - 131} = \sqrt{6.25} = 2.5$ .
Circle 2 ( $C_2$ ): Center $O_2 = (3, 3)$ , Radius $r_2 = \sqrt{3^2 + 3^2 - (-7)} = \sqrt{9 + 9 + 7} = \sqrt{25} = 5$ .
Distance between centers ( $d$ ):

$d = \sqrt{(9-3)^2 + (7.5-3)^2} = \sqrt{6^2 + 4.5^2} = \sqrt{36 + 20.25} = \sqrt{56.25} = 7.5$
Comparison: Yahan $d = 7.5$ hai aur $r_1 + r_2 = 2.5 + 5 = 7.5$ .

Kyuki $d = r_1 + r_2$ , dono circles ek dusre ko bahar se touch (touch externally) kar rahe hain.

Jab do circles externally touch karte hain, toh unke 3 common tangents hote hain.

Correct Option: (3)