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 [2023]

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Solution

Q.
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 [2023]

1 2

2 3

3 4

4 1

Ans.
(2)

$C_{1} : x^{2} + y^{2} - 18 x - 15 y + 131 = 0$

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

We know that for equation of circle

$x^{2} + y^{2} + 2 g x + 2 f y + C = 0$

Centre is $(- g, - f)$ and radius $= \sqrt{g^{2} + f^{2} - C}$

For $C_{1},$ $Centre (A) = (9, \frac{15}{2})$

$and radius (r_{1}) = \sqrt{81 + \frac{225}{4} - 131} = \sqrt{\frac{25}{4}} = \frac{5}{2}$

For $C_{2},$ $Centre (B) = (3, 3)$

$and radius (r_{2}) = \sqrt{9 + 9 + 7} = 5$

and $A B = \sqrt{{(6)}^{2} + {(\frac{9}{2})}^{2}} = \sqrt{36 + \frac{81}{4}} = \sqrt{\frac{225}{4}} = \frac{15}{2} = r_{1} + r_{2}$

$∴$ $There are 3 common tangents$

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