The number of common tangents to the circles | vedclass

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Question

(A) For the first circle $x^2+y^2-18x-15y+131=0$,the center $C_1 = (9, 7.5)$ and radius $r_1 = \sqrt{9^2 + 7.5^2 - 131} = \sqrt{81 + 56.25 - 131} = \s

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) For the first circle $x^2+y^2-18x-15y+131=0$,the center $C_1 = (9, 7.5)$ and radius $r_1 = \sqrt{9^2 + 7.5^2 - 131} = \sqrt{81 + 56.25 - 131} = \sqrt{6.25} = 2.5 = \frac{5}{2}$.For the second circle $x^2+y^2-6x-6y-7=0$,the center $C_2 = (3, 3)$ and radius $r_2 = \sqrt{3^2 + 3^2 - (-7)} = \sqrt{9 + 9 + 7} = \sqrt{25} = 5$.The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(9-3)^2 + (7.5-3)^2} = \sqrt{6^2 + 4.5^2} = \sqrt{36 + 20.25} = \sqrt{56.25} = 7.5 = \frac{15}{2}$.Since $r_1 + r_2 = 2.5 + 5 = 7.5$,we have $d = r_1 + r_2$.Because the distance between the centers is equal to the sum of the radii,the circles touch each other externally.Therefore,the number of common tangents is $3$.

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:

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