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

Sep 20, 2015

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

$\left(x , y\right) = \left(\frac{5 \pm \sqrt{1215}}{20} , - \frac{135 \pm \sqrt{1215}}{60}\right)$

(with matching signs for $\pm$)

#### Explanation:

Subtract the second equation from the first to get:

$6 y + 2 x + 13 = 0$

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

$6 y = - \left(2 x + 13\right)$

Divide both side by $6$ to get:

$y = - \frac{2 x + 13}{6}$

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

$0 = {x}^{2} + {\left(2 x + 13\right)}^{2} / 36 - \left(2 x + 13\right) + 5$

$= {x}^{2} + {\left(2 x + 13\right)}^{2} / 36 - 2 x - 8$

Multiply through by $36$ to get:

$0 = 36 {x}^{2} + {\left(2 x + 13\right)}^{2} - 72 x - 288$

$= 36 {x}^{2} + 4 {x}^{2} + 52 x + 169 - 72 x - 288$

$= 40 {x}^{2} - 20 x - 119$

Solve this using the quadratic formula to get:

$x = \frac{20 \pm \sqrt{{20}^{2} - \left(4 \times 40 \times - 119\right)}}{2 \times 40}$

$= \frac{20 \pm \sqrt{400 + 19040}}{80}$

$= \frac{20 \pm \sqrt{19440}}{80}$

$= \frac{5 \pm \sqrt{1215}}{20}$

So:

$y = - \frac{2 \left(\frac{5 \pm \sqrt{1215}}{20}\right) + 13}{6}$

$= - \frac{20 \left(\frac{5 \pm \sqrt{1215}}{20}\right) + 130}{60}$

$= - \frac{5 \pm \sqrt{1215} + 130}{60}$

$= - \frac{135 \pm \sqrt{1215}}{60}$

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