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

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Answer 1 · 2015-09-20T19:00:41

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

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

(with matching signs for $+-$ )

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]}

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