by **phw** » Fri Jun 08, 2018 12:25 pm

You need to prove two things. Namely, that

[tex](X^2+Y^2+A_1X+B_1Y+C_1) + k(X^2+Y^2+A_2X+B_2Y+C_2 )=0[/tex]

(1) contains the points common to the earlier given circles, and

(2) is a circle.

For (1), suppose [tex](x_0, y_0)[/tex] is a point of intersection. Since it is on the first circle, then [tex]x_0^2+y_0^2+A_1x_0+B_1y_0+C_1 = 0[/tex] and since it is also on the second, then [tex]x_0^2+y_0^2+A_2x_0+B_2y_0+C_2 = 0[/tex]. So plugging [tex](x_0, y_0)[/tex] into the big equation simplifies to [tex]0 + k\cdot0 = 0[/tex] which is obviously an identity for any [tex]k[/tex]. Conclude that it does indeed contain the required points, but is it a circle?

For (2), expand and simplify. You should get [tex](1+k)X^2 + (1+k)Y^2 + [\text{lower degree terms}] = 0[/tex]. Recall that the nature of a conic is determined by the second degree terms, and that when X^2 and Y^2 have the same coefficient (and the XY term is absent), the conic is a circle. Except of course that if [tex]k=-1[/tex], all the second degree terms vanish and you get a degenerate conic (probably the line containing the point of intersection).

But if [tex]k\ne-1[/tex], then the equation describes a circle and it contains at least the common points of the other two circles.