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?

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

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

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

To show that the given equation represents a circle passing through the points of intersection of C1 and C2, we need to determine the center and radius of the circle.

Step 1: Find the points of intersection of C1 and C2
To find the points of intersection, we can set the two given equations equal to each other:

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

Rearranging and simplifying the equation, we get:

x^2 - y^2 - 6y = 0

Factorizing, we have:

(x - y)(x + y + 6) = 0

Setting each factor equal to zero, we get two equations:

x - y = 0 (Equation 1)
x + y + 6 = 0 (Equation 2)

Solving these equations simultaneously gives us the points of intersection.

Step 2: Determine the center and radius of the circle
To determine the center and radius of the circle passing through the points of intersection, we can use the general equation of a circle:

(x - h)^2 + (y - k)^2 = r^2

We substitute the coordinates of one of the points of intersection into this equation and solve for h, k, and r.

For example, let's use Equation 1 to find the center and radius:

x - y = 0

Solving for x, we get:

x = y

Substituting x = y into C1 gives:

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

Combining like terms:

2x^2 + 4x - 3 = 0

Using the quadratic formula, we find that x = -1 and x = 1/2.

So, the two points of intersection are (-1, -1) and (1/2, 1/2).

Substituting the coordinates of one of the points into the general equation of a circle, we can solve for h, k, and r:

(-1 - h)^2 + (-1 - k)^2 = r^2

Simplifying, we get:

(h + 1)^2 + (k + 1)^2 = r^2

Solving for h, k, and r will give us the center and radius of the circle.

(a) To find the particular value of λ for which the circle passes through the origin (0, 0), we substitute x = 0 and y = 0 into the equation:

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

This gives us:

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

Simplifying, we get:

-3 + 3λ = 0

Solving for λ, we find:

λ = 1

So, for λ = 1, the circle passes through the origin (0, 0).

(b) To determine what happens when λ = -1/2, we substitute this value into the given equation:

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

Simplifying, we get:

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

Combining like terms, we have:

2x - 3/2 + 3y = 0

This equation represents a line and not a circle. Therefore, when λ = -1/2, the equation does not represent a circle.