The number of common tangents to the circles | vedclass

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(B) For the first circle $x^2+y^2-4x-6y-12=0$: Center $C_1 = (2, 3)$, Radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$. For the sec

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$3$

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(B) For the first circle $x^2+y^2-4x-6y-12=0$: Center $C_1 = (2, 3)$, Radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$. For the second circle $x^2+y^2+6x+18y+26=0$: Center $C_2 = (-3, -9)$, Radius $r_2 = \sqrt{(-3)^2+(-9)^2-26} = \sqrt{9+81-26} = \sqrt{64} = 8$. The distance between centers $C_1C_2 = \sqrt{(2 - (-3))^2 + (3 - (-9))^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$. Since $r_1 + r_2 = 5 + 8 = 13$, we have $C_1C_2 = r_1 + r_2$. This implies that the two circles touch each other externally. When two circles touch each other externally, the number of common tangents is $3$.

Column $I$	Column $II$
$(a)$ $XX$-$XO$ method of sex determination	$(i)$ Turner's syndrome
$(b)$ $XX$-$XY$ method of sex determination	$(ii)$ Female heterogametic
$(c)$ Karyotype-$45$	$(iii)$ Grasshopper
$(d)$ $ZW$-$ZZ$ method of sex determination	$(iv)$ Female homogametic

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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