Question

In our math class, we didn't really go through alot on The circle so I still have some problem about the question assigned to me :


Two circle C1 and C2 have equations
x^2 + y^2 +4x - 3 = 0 and


2x^2 + 2y^2 +4x -6y + 3 =0


respectively. Show that, for all values of λ, the equation

x^2 + y^2 +4x - 3 + λ(2x^2 + 2y^2 +4x -6y + 3 = 0)

represents a circle passing through the points of intersection of C1 and C2
(a) For what particular value of λ does this circle pass through origin?

(b) What hsppens when λ = -1/2?

Answer 1

Solving the system of equations

x^2 + y^2 +4x - 3 = 0
2x^2 + 2y^2 +4x -6y + 3 =0

Is the same as solving the system

x^2 + y^2 +4x - 3 = 0
λ(2x^2 + 2y^2 +4x -6y + 3) = 0

The equation

x^2 + y^2 +4x - 3 + λ(2x^2 + 2y^2 +4x -6y + 3) = 0

can be written as (*)

(1+2λ)x^2 + 4(1+2λ)x + (1+2λ)y^2 - 6λy = 3(1-λ)

(1+2λ)(x^2+4x+4) + (1+2λ)(y^2-(6λ/(1+2λ) y + (3λ/(1+2λ))^2) = 3(1-λ)+4(1+2λ)+(9λ^2)/(1+2λ)

(1+2λ)(x+2)^2 + (1+2λ)(y - 3λ/(1+2λ))^2 = (19λ^2+19λ+7)/(1+2λ)

(x+2)^2 + (y - 3λ/(1+2λ))^2 = (19λ^2+19λ+7)/(1+2λ)^2

This circle passes through (0,0) when

4 + 9λ^2/(1+2λ)^2 = (19λ^2+19λ+7)/(1+2λ)^2

4(1+2λ)^2 + 9λ^2 = 19λ^2+19λ+7
6λ^2-3λ-3 = 0
(λ-1)(2λ+1) = 0
So, the circle passes through (0,0) when λ=1

Why not when λ = -1/2 ? Scroll up to look at the equation (*). When λ = -1/2 the equation degenerates into a line:

- 6λy = 3(1-λ)
y = 3/2

Answer 2

When ?=1, we get the circle

(x+2)^2 + (y-1)^2 = 5

The three circles can be seen at

http://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+y%5E2+%2B4x+-+3+%3D+0,+++2x%5E2+%2B+2y%5E2+%2B4x+-6y+%2B+3+%3D0,+(x%2B2)%5E2+%2B+(y-1)%5E2+%3D+5,+y%3D0,+x%3D0

Answer 3

My bad. I should have said that when ?=1 we get the equation

x^2 + y^2 +4x - 3 + 1(2x^2 + 2y^2 +4x -6y + 3) = 0

3x^2+8x + 3y^2-6y = 0

(x + 4/3)^2 + (y-1)^2 = 25/9

Now the three circles can all be seen to intersect at the same two points:

http://www.wolframalpha.com/input/?i=plot+x%5E2+%2B+y%5E2+%2B4x+-+3+%3D+0,+++2x%5E2+%2B+2y%5E2+%2B4x+-6y+%2B+3+%3D0,+(x+%2B+4%2F3)%5E2+%2B+(y-1)%5E2+%3D+25%2F9,+y%3D0,x%3D0