How to find the intersections of a circle and a line?

[Casio1]

I want to find the coordinates of any points at which the circle x^2 + y^2 + 6x - 8y + 20 = 0

I have an equation y = -2x - 2

I have tried the following but am struggling with the final understanding.

(x + 3)^2 + (y - 4)^2 = 5

(x + 3)^2 = x^2 + 6x +9

(y - 4)^2 = 4x^2 + 24x + 36

combining and cancelling gives

5x^2 + 30x + 39 = 0

I now require to find the factors, which I am struggling with, I am thinking

x^2 + 10x + 13

In my head I have these numbers running round, factors 3, 6 and 5. I know there is a method to working this out like 3 x 13 = 39, and 6 x 5 = 30 etc, but I just can't seem to grasp it?

Any help appreciated

Thanks(Wondering)

 

[Ackbach]

I want to find the coordinates of any points at which the circle x^2 + y^2 + 6x - 8y + 20 = 0

I have an equation y = -2x - 2

I have tried the following but am struggling with the final understanding.

(x + 3)^2 + (y - 4)^2 = 5

(x + 3)^2 = x^2 + 6x +9

(y - 4)^2 = 4x^2 + 24x + 36

combining and cancelling gives

5x^2 + 30x + 39 = 0

I now require to find the factors, which I am struggling with, I am thinking

x^2 + 10x + 13

In my head I have these numbers running round, factors 3, 6 and 5. I know there is a method to working this out like 3 x 13 = 39, and 6 x 5 = 30 etc, but I just can't seem to grasp it?

Any help appreciated

Thanks(Wondering)

I'm a little unsure what you're trying to do. Are you trying to find the intersection of the circle with the straight line?

 

[soroban]

Hello, Casio!

I assume you want the intersections of the circle and the line . . .

\begin{array}{cc}x^2 + y^2 + 6x - 8y + 20 \:=\: 0 \\ y \:=\: -2x - 2 \end{array}

Why not substitute directly?

\begin{array}{cc}x^2 + (-2x-2)^2 + 6x - 8(-2x-2) + 20 \:=\:0 \\ x^2 + 4x^2 + 8x + 4 + 6x + 16x + 16 + 20 \:=\:0 \\ 5x^2 + 30x + 40 \:=\:0 \\ x^2 + 6x + 8 \:=\:0 \\ (x+2)(x+4) \:=\:0 \end{array}

\begin{Bmatrix}x = -2 \\ x = -4 \end{Bmatrix} . \Rightarrow . \begin{Bmatrix} y = 2 \\ y = 6 \end{Bmatrix}Intersections: .(-2,\,2),\;(-4,\,6)