How do you find the center and radius of the circle $x^{2} + y^{2} - 10 x + 8 y + 32 = 0$ $x^2 + y^2 − 10x + 8y + 32 = 0$ ?

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How do you find the center and radius of the circle $x^{2} + y^{2} - 10 x + 8 y + 32 = 0$ $x^2 + y^2 − 10x + 8y + 32 = 0$ ?

1 Answer

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Answer 1 · 2016-10-16T18:42:15

The center is $(5, - 4)$ $(5,-4)$ and the radius is 3

Explanation:

$x^{2} - 10 x + y^{2} + 8 y + 32 = 0$ $x^2-10x+y^2+8y+32=0$
Completing the squares
$x^{2} - 10 x + 25 + y^{2} + 8 y + 16 = - 32 + 25 + 16$ $x^2-10x+25+y^2+8y+16=-32+25+16$
${(x - 5)}^{2} + {(y + 4)}^{2} = 9 = 3^{2}$ $(x-5)^2+(y+4)^2=9=3^2$
So the center is $(5, - 4)$ $(5,-4)$ and the radius is 3
See the graph below
graph{x^2-10x+y^2+8y+32=0 [-20, 20, -10, 10]}

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How do you find the center and radius of the circle $x^{2} + y^{2} - 10 x + 8 y + 32 = 0$ $x^2 + y^2 − 10x + 8y + 32 = 0$ ?

1 Answer

Explanation:

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How do you find the center and radius of the circle x2+y2−10x+8y+32=0x^2 + y^2 − 10x + 8y + 32 = 0?

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Explanation:

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How do you find the center and radius of the circle $x^{2} + y^{2} - 10 x + 8 y + 32 = 0$ $x^2 + y^2 − 10x + 8y + 32 = 0$ ?