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.

Two circles, with centres $C_1,C_2$ and radii $r_1,r_2$, can sit relative to each other in exactly five ways, distinguished entirely by comparing the distance $d(C_1C_2)$ between their centres against $r_1$ and $r_2$: touching externally ($d(C_1C_2)=r_1+r_2$, three common tangents), touching internally ($d(C_1C_2)=|r_1-r_2|$, one common tangent), disjoint ($r_1+r_2<d(C_1C_2)$, four common tangents), intersecting in two points (the line through those two points is the common chord, also called the radical axis; two common tangents), and concentric (same centre, different radii; no common tangent). When two circles touch (externally or internally), the point of contact lies on the segment joining their centres, and can be found by dividing that segment in the ratio $r_1:r_2$ (internally for external contact, externally for internal contact); the common tangent line at that point of contact can then be found either as the tangent to either circle there, or — more quickly — by subtracting the two circles' equations, since the $x^2,y^2$ terms cancel and leave exactly the linear equation of the radical axis, which coincides with the common tangent at a point of tang …