Question

How do you solve `x^2+y^2+6y+5=0` and `x^2+y^2-2x-8=0`?

Answer 1

Resolve to a quadratic in `x` and solve to find:

`(x, y) = ((5+-sqrt(1215))/20, -(135+-sqrt(1215))/60)`

(with matching signs for `+-`)

Explanation:

Subtract the second equation from the first to get:

`6y + 2x + 13 = 0`

Subtract `2x + 13` from both sides to get:

`6y = -(2x+13)`

Divide both side by `6` to get:

`y = -(2x+13)/6`

Substitute this expression for `y` in the first equation to get:

`0 = x^2 + (2x+13)^2/36 -(2x+13)+5`

`= x^2 + (2x+13)^2/36-2x-8`

Multiply through by `36` to get:

`0 = 36x^2 + (2x+13)^2 -72x - 288`

`=36x^2 + 4x^2+52x+169-72x-288`

`=40x^2-20x-119`

Solve this using the quadratic formula to get:

`x = (20 +-sqrt(20^2-(4xx40xx-119)))/(2xx40)`

`= (20 +- sqrt(400+19040))/80`

`=(20 +- sqrt(19440))/80`

`=(5+-sqrt(1215))/20`

So:

`y = -(2((5+-sqrt(1215))/20)+13)/6`

`=-(20((5+-sqrt(1215))/20)+130)/60`

`=-(5+-sqrt(1215)+130)/60`

`=-(135+-sqrt(1215))/60`

graph{(x^2+y^2+6y+5)(x^2+y^2-2x-8) = 0 [-9.755, 10.245, -6.04, 3.96]}