The general form of the equation of a circle is x2+y2−4x−8y−5=0. What are the coordinates of the center of the circle?

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The general form of the equation of a circle is x2+y2−4x−8y−5=0. What are the coordinates of the center of the circle?

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Answer 1 · 2017-05-11T14:10:03

The center of the circle is at $(2, 4)$ $(2,4)$ and the circle has a radius of $5$ $5$ .

Explanation:

Formatted question: $x^{2} + y^{2} - 4 x - 8 y - 5 = 0$ $x^2+y^2-4x-8y-5=0$

Note: The standard form a circle is ${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$ , where $(h, k)$ $(h,k)$ is the center of the circle and $r$ $r$ is the radius of the circle.

To find the standard form of the circle, we need to complete the square and move the constants to the right side:
$x^{2} + y^{2} - 4 x - 8 y = 5$ $x^2+y^2-4x-8y=5$
${(x - 2)}^{2} - 4 + {(y - 4)}^{2} - 16 = 5$ $(x-2)^2-4+(y-4)^2-16=5$
${(x - 2)}^{2} + {(y - 4)}^{2} = 25$ $(x-2)^2+(y-4)^2=25$

Therefore, the center of the circle is at $(2, 4)$ $(2,4)$ and the circle has a radius of $5$ $5$ .

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The general form of the equation of a circle is x2+y2−4x−8y−5=0. What are the coordinates of the center of the circle?

1 Answer

Explanation:

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