What is the length of the radius and the coordinates of the center of the circle defined by the equation ${(x + 7)}^{2} + {(y - 3)}^{2} = 121$ $(x+7)^2+(y-3)^2=121$ ?

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What is the length of the radius and the coordinates of the center of the circle defined by the equation ${(x + 7)}^{2} + {(y - 3)}^{2} = 121$ $(x+7)^2+(y-3)^2=121$ ?

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Answer 1 · 2017-07-06T14:54:20

In

${(x - p)}^{2} + {(y - q)}^{2} = r^{2}$ $(x - p)^2 + (y - q)^2 = r^2$

The radius is given by $r$ $r$ and the centre by $(p, q)$ $(p, q)$ . Hence, our centre is $(- 7, 3)$ $(-7, 3)$ and the radius is $11$ $11$ . The graph of the relation confirms.

graph{(x+ 7)^2 + (y - 3)^2 = 121 [-10, 10, -5, 5]}

Hopefully this helps!

Answer 2 · 2017-07-06T14:58:43

Radius is $11$ $11$ and coordinates of center are $(- 7, 3)$ $(-7,3)$

Explanation:

Equation of a circle whose center is $(h, k)$ $(h,k)$ and radius is $r$ $r$ is

${(x - h)}^{2} + {(y - k)}^{2} = r^{2}$ $(x-h)^2+(y-k)^2=r^2$

As the equation ${(x + 7)}^{2} + {(y - 3)}^{2} = 121$ $(x+7)^2+(y-3)^2=121$ can be written as

${(x - (- 7))}^{2} + {(y - 3)}^{2} = 11^{2}$ $(x-(-7))^2+(y-3)^2=11^2$

Radius is $11$ $11$ and coordinates of center are $(- 7, 3)$ $(-7,3)$

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What is the length of the radius and the coordinates of the center of the circle defined by the equation ${(x + 7)}^{2} + {(y - 3)}^{2} = 121$ $(x+7)^2+(y-3)^2=121$ ?

2 Answers

Explanation:

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What is the length of the radius and the coordinates of the center of the circle defined by the equation (x+7)2+(y−3)2=121(x+7)^2+(y-3)^2=121?

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Explanation:

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What is the length of the radius and the coordinates of the center of the circle defined by the equation ${(x + 7)}^{2} + {(y - 3)}^{2} = 121$ $(x+7)^2+(y-3)^2=121$ ?