Q.Show that the circles $x^2+y^2-8x-2y+8=0$ and $x^2+y^2-2x+6y+6=0$ touch each other and find the point of contact.

Compute both centres and radii; the circles touch externally because the distance between centres equals the sum of radii, and the point of contact divides that segment in the ratio $r_1:r_2$.

Circle 1: $x^2+y^2-8x-2y+8=0$: centre $C_1=(4,1)$, $r_1=\sqrt{16+1-8}=3$

Circle 2: $x^2+y^2-2x+6y+6=0$: centre $C_2=(1,-3)$, $r_2=\sqrt{1+9-6}=2$

Distance between centres:

$C_1C_2 = \sqrt{(4-1)^2+(1-(-3))^2}=\sqrt{9+16}=5$

Since $C_1C_2 = r_1+r_2 = 3+2=5$, the circles touch each other externally.

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