Examine if the two circles $x^{2} + y^{2} - 2 x - 4 y = 0$ and $x^{2} + y^{2} - 8 y - 4 = 0$ touch each other externally or internally

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Solution

Given circles are $x^{2} + y^{2} - 2 x - 4 y = 0$ .................(i)
and $x^{2} + y^{2} - 8 y - 4 = 0$ ...............(ii)
Let A and B be the centres and $r_{1}$ and $r_{2}$ the radii of circles (i) and (ii) respectively then
$A = (1, 2), B \equiv (0, 4), r_{1} = \sqrt{5}, r_{2} = 2 \sqrt{5}$
Now $A B = \sqrt{(1 - 0)^{2} + (2 - 4)^{2}} = \sqrt{5} a n d r_{1} + r_{2} = 3 \sqrt{5}, | r_{1} - r_{2} | = \sqrt{5}$
Thus AB = $| r_{1} - r_{2} |$ hence the two circles touch each other internally

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