Given the equation $2 x^{2} + 2 y^{2} + 24 x - 81 = 0 ?$ $2x^2+2y^2+24x-81=0?$

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Find the center and radius of a circle

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Answer 1 · 2017-02-24T09:20:57

Center is $(- 6, 0)$ $(-6,0)$ and radius is $8.746$ $8.746$

$2 x^{2} + 2 y^{2} + 24 x - 81 = 0$ $2x^2+2y^2+24x-81=0$ can be written as (dividing each term by $2$ $2$ )

$x^{2} + y^{2} + 12 x - \frac{81}{2} = 0$ $x^2+y^2+12x-81/2=0$

or $x^{2} + 2 \times x \times 6 + 6^{2} + y^{2} - 6^{2} - \frac{81}{2} = 0$ $x^2+2xx x xx6+6^2+y^2-6^2-81/2=0$

or ${(x + 6)}^{2} + y^{2} = \frac{81}{2} + 36 = \frac{153}{2}$ $(x+6)^2+y^2=81/2+36=153/2$

or ${(x - (- 6))}^{2} + {(y - 0)}^{2} = {(\sqrt{\frac{153}{2}})}^{2}$ $(x-(-6))^2+(y-0)^2=(sqrt(153/2))^2$

which is locus of a point which moves so that

its distance from $(- 6, 0)$ $(-6,0)$ is always $\sqrt{\frac{153}{2}} = 8.746$ $sqrt(153/2)=8.746$

Hence, center is $(- 6, 0)$ $(-6,0)$ and radius is $8.746$ $8.746$
graph{2x^2+2y^2+24x-81=0 [-20, 20, -10, 10]}