The number of common tangents to the circles | vedclass

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(D) For the first circle ${x^2} + {y^2} - 4x - 6y - 12 = 0$,the center $c_{1} = (2, 3)$ and radius $r_{1} = \sqrt{2^2 + 3^2 - (-12)} = \sqrt{4 + 9 + 1

Vedclass · Accepted Answer

$3$

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(D) For the first circle ${x^2} + {y^2} - 4x - 6y - 12 = 0$,the center $c_{1} = (2, 3)$ and radius $r_{1} = \sqrt{2^2 + 3^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$.For the second circle ${x^2} + {y^2} + 6x + 18y + 26 = 0$,the center $c_{2} = (-3, -9)$ and radius $r_{2} = \sqrt{(-3)^2 + (-9)^2 - 26} = \sqrt{9 + 81 - 26} = \sqrt{64} = 8$.The distance between the centers $c_{1}c_{2} = \sqrt{(2 - (-3))^2 + (3 - (-9))^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.Since $c_{1}c_{2} = r_{1} + r_{2} = 5 + 8 = 13$,the two circles touch each other externally.When two circles touch each other externally,the number of common tangents is $3$.

The number of common tangents to the circles ${x^2} + {y^2} - 4x - 6y - 12 = 0$ and ${x^2} + {y^2} + 6x + 18y + 26 = 0$ is

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