The number of common tangent(s) to the circle | vedclass

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Question

Let $S_{1}: x^{2}+y^{2}+2 x+8 y-23=0$ and $S_{2}: x^{2}+y^{2}-4 x-10 y+19=0$

$\Rightarrow C_{1}=(-1,-4), r_{1}=\sqrt{1+16+23}=2 \sqrt{10}, C_{2}=(2

Vedclass · Accepted Answer

$3$

Vedclass · Answer

Let $S_{1}: x^{2}+y^{2}+2 x+8 y-23=0$ and $S_{2}: x^{2}+y^{2}-4 x-10 y+19=0$

$\Rightarrow C_{1}=(-1,-4), r_{1}=\sqrt{1+16+23}=2 \sqrt{10}, C_{2}=(2,5), r_{2}=\sqrt{4+25-19}=\sqrt{10}$

Now $C_{1} C_{2}=\sqrt{3^{2}+9^{2}}=3 \sqrt{10}$

Clearly $C_{1} C_{2}=r_{1}+r_{2}$ i.e both the circles touch each other externally.

Hence there will be 3 common tangent.

The number of common tangent$(s)$ to the circles $x^2 + y^2 + 2x + 8y - 23 = 0$ and $x^2 + y^2 - 4x - 10y + 19 = 0$ is :

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