# Intersecting circle formula proof(help)

### Intersecting circle formula proof(help)

let L1 and L2 be two intersecting circles determined by the equations X^2+Y^2+A1X+B1Y+C1=0 and X^2+Y^2+A2X+B2Y+C2=0
for any number k different that -1,show that

let L1 and L2 be two intersecting circles determined by the equations X^2+Y^2+A1X+B1Y+C1=0 and X^2+Y^2+A2X+B2Y+C2=0
for any number k different that -1,show that

X^2+Y^2+A1X+B1Y+C1+k(X^2+Y^2+A2X+B2Y+C2 )=0
is the equation of a circle through the intersection points L1 and L2

Guest

### Re: intersecting circle formula proof(help)

Guest wrote:let L1 and L2 be two intersecting circles determined by the equations X^2+Y^2+A1X+B1Y+C1=0 and X^2+Y^2+A2X+B2Y+C2=0
for any number k different that -1,show that

let L1 and L2 be two intersecting circles determined by the equations X^2+Y^2+A1X+B1Y+C1=0 and X^2+Y^2+A2X+B2Y+C2=0
for any number k different that -1,show that

X^2+Y^2+A1X+B1Y+C1+k(X^2+Y^2+A2X+B2Y+C2 )=0
is the equation of a circle through the intersection points L1 and L2

from ayres,calculus 5 ed question 24,chapter 4,i need your help guys

Guest

### Re: intersecting circle formula proof(help)

You need to prove two things. Namely, that

$(X^2+Y^2+A_1X+B_1Y+C_1) + k(X^2+Y^2+A_2X+B_2Y+C_2 )=0$

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

(2) is a circle.

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

For (2), expand and simplify. You should get $(1+k)X^2 + (1+k)Y^2 + [\text{lower degree terms}] = 0$. 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 $k=-1$, all the second degree terms vanish and you get a degenerate conic (probably the line containing the point of intersection).

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

phw

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