# Intersections 3

Find the intersections of the circles
x2 + y2 + 6 x - 10 y + 9 = 0 and
x2 + y2 + 18 x + 4 y + 21 = 0

Correct result:

x1 =  -1.334
y1 =  0.286
x2 =  -7.914
y2 =  5.926

#### Solution:

$x^2 + y^2 + 6 \ x - 10 \ y + 9=0 \ \\ x^2 + y^2 + 18 \ x + 4 \ y + 21=0 \ \\ \ \\ e_{2}-e_{1}: \ \\ (18-6)x + (4+10)y + (21-9)=0 \ \\ y=-6/7 (x + 1) \ \\ \ \\ \ \\ x^2 + (-6/7 (x + 1))^2 + 6 x - 10 (-6/7 (x + 1)) + 9=0 \ \\ \ \\ x^2 + (-6/7 (x + 1))^2 + 6 \ x - 10 (-6/7 (x + 1)) + 9=0 \ \\ 1.734693877551x^2 +16.041x +18.306=0 \ \\ \ \\ a=1.734693877551; b=16.041; c=18.306 \ \\ D=b^2 - 4ac=16.041^2 - 4\cdot 1.734693877551 \cdot 18.306=130.2857142857 \ \\ D>0 \ \\ \ \\ x_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -16.04 \pm \sqrt{ 130.29 } }{ 3.469387755102 } \ \\ x_{1,2}=-4.62352941 \pm 3.2899974232486 \ \\ x_{1}=-1.3335319885161=-1.334 \ \\ x_{2}=-7.9135268350134 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 1.734693877551 (x +1.3335319885161) (x +7.9135268350134)=0$

Checkout calculation with our calculator of quadratic equations.

$y_{1}=-6/7 \cdot \ (x_{1} + 1)=-6/7 \cdot \ ((-1.3335) + 1)=\dfrac{ 501 }{ 1750 }=0.286$
$y_{2}=-6/7 \cdot \ (x_{2} + 1)=-6/7 \cdot \ ((-7.9135) + 1)=5.926$

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