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.