The number of common tangents to the circles | vedclass

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(D) The given equations of the circles are:$C_{1}: x^{2}+y^{2}=4$,with center $O_{1}=(0,0)$ and radius $r_{1}=2$.$C_{2}: x^{2}+y^{2}-6x-8y-24=0$,with

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$1$

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(D) The given equations of the circles are:$C_{1}: x^{2}+y^{2}=4$,with center $O_{1}=(0,0)$ and radius $r_{1}=2$.$C_{2}: x^{2}+y^{2}-6x-8y-24=0$,with center $O_{2}=(3,4)$ and radius $r_{2}=\sqrt{3^{2}+4^{2}-(-24)}=\sqrt{9+16+24}=\sqrt{49}=7$.Now,calculate the distance between the centers $O_{1}$ and $O_{2}$:$d = \sqrt{(3-0)^{2}+(4-0)^{2}} = \sqrt{9+16} = 5$.Compare the distance $d$ with the sum and difference of the radii:$r_{1}+r_{2} = 2+7 = 9$.$|r_{1}-r_{2}| = |2-7| = 5$.Since $d = |r_{1}-r_{2}|$,the circles touch each other internally.Therefore,the number of common tangents is $1$.

The number of common tangents to the circles $x^{2}+y^{2}=4$ and $x^{2}+y^{2}-6x-8y-24=0$ is

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