The minimum distance between any two points $P_{1}$ and $P_{2}$ while considering point $P_{1}$ on one circle and point $P_{2}$ on the other circle for the given circle's equations $x^{2} + y^{2} - 10 x - 10 y + 41 = 0, x^{2} + y^{2} - 24 x - 10 y + 160 = 0$

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Solution

Given :
$S_{1} : x^{2} + y^{2} - 10 x - 10 y + 41 = 0 S_{2} : x^{2} + y^{2} - 24 x - 10 y + 160 = 0$
Centre and radius of the circles are
$C_{1} = (5, 5), r_{1} = 3 C_{2} = (12, 5), r_{2} = 3 C_{1} C_{2} = 7$

So minimum distance between $P_{1}$ and $P_{2}$
$= C_{1} C_{2} - (r_{1} + r_{2}) = 7 - (3 + 3) = 1$

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