The number of common tangents to the circles | vedclass

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Question

(C) The equation of the first circle is ${x^2 + y^2 - 4x - 6y - 12 = 0}$. Its centre ${C_1}$ is $(2, 3)$ and radius ${r_1} = \sqrt{2^2 + 3^2 - (-12)}

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The equation of the first circle is ${x^2 + y^2 - 4x - 6y - 12 = 0}$. Its centre ${C_1}$ is $(2, 3)$ and radius ${r_1} = \sqrt{2^2 + 3^2 - (-12)} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$.The equation of the second circle is ${x^2 + y^2 + 6x + 18y + 26 = 0}$. Its centre ${C_2}$ is $(-3, -9)$ and radius ${r_2} = \sqrt{(-3)^2 + (-9)^2 - 26} = \sqrt{9 + 81 - 26} = \sqrt{64} = 8$.The distance between the centres ${C_1}$ and ${C_2}$ is ${d = \sqrt{(2 - (-3))^2 + (3 - (-9))^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13}$.Since ${r_1 + r_2 = 5 + 8 = 13}$,which is equal to the distance between the centres ${d}$,the circles touch each other externally.For two circles touching 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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