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 answer:

x₁ =  -1.3335
y₁ =  0.2859
x₂ =  -7.9135
y₂ =  5.9259

Step-by-step explanation:

$$\begin{gathered} x^2+y^2+6x-10y+9=0 \\ {x}^2+y^2+18x+4y+21=0 \\ {e}_2-e_1: \\ (18-6)x+(4+10)y+(21-9)=0 \\ y=-6\mathrm{/}7(x+1) \\ {x}^2+(-6\mathrm{/}7(x+1){)}^2+6x-10(-6\mathrm{/}7(x+1))+9=0 \\ {x}^2+(-6\mathrm{/}7(x+1){)}^2+6x-10(-6\mathrm{/}7(x+1))+9=0 \\ 1.73469387755x^2+16.041x+18.306=0 \\ a=1.73469387755;b=16.041;c=18.306 \\ D=b^2-4ac=16.041^2-4\cdot 1.73469387755\cdot 18.306=130.285714286 \\ D>0 \\ {x}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{-16.04\pm \sqrt{130.29}}{3.4693877551}} \\ {x}_{1,2}=-4.62352941\pm 3.28999742325 \\ {x}_1=-1.33353198852=-1.3335 \\ {x}_2=-7.91352683501 \\ \text{ Factored form of the equation: } \\ 1.73469387755(x+1.33353198852)(x+7.91352683501)=0 \end{gathered}$$

Our quadratic equation calculator calculates it.

$y_1=-6\mathrm{/}7\cdot (x_1+1)=-6\mathrm{/}7\cdot ((-1.3335)+1)=0.2859$

$y_2=-6\mathrm{/}7\cdot (x_2+1)=-6\mathrm{/}7\cdot ((-7.9135)+1)=5.9259$

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