The centre of the smallest circle touching th | vedclass

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Question

let

$S_{1}=x^{2}+y^{2}-2 y-3=0$

$C_{1}(0,1), r_{1}=2$

$S_{2}: x^{2}+y^{2}-8 x-18 y+93=0$

$C_{2}(4,9), r_{2}=2$

Hence, for smallest circ

Vedclass · Accepted Answer

$(2 , 5)$

Vedclass · Answer

let

$S_{1}=x^{2}+y^{2}-2 y-3=0$

$C_{1}(0,1), r_{1}=2$

$S_{2}: x^{2}+y^{2}-8 x-18 y+93=0$

$C_{2}(4,9), r_{2}=2$

Hence, for smallest circle ${ }^{\prime} C_{3}^{\prime}$ centre is midpoint of ${ }^{\prime} C_{1}^{\prime},{ }^{\prime} C_{2}^{\prime}$ $	herefore C_{3}\left(\frac{0+4}{2}, \frac{9+1}{2}ight)$

$C_{3}(2,5)$

The centre of the smallest circle touching the circles $x^2 + y^2- 2y - 3 = 0$ and $x^2+ y^2 - 8x - 18y + 93 = 0$ is :

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