How do you find the center and radius of the circle given $x^{2} + {(y + 2)}^{2} = 4$ $x^2+(y+2)^2=4$ ?

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Answer 1 · 2016-11-24T17:06:17

Please see the explanation.

The standard form for the equation of a circle is:

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

where $(x, y)$ $(x, y)$ is any point on the circle, $(h, k)$ $(h, k)$ is the center, and r is the radius.

I suggest that you modify any given equation into this form and then report the easily observed values.

Change the + 2 to - -2:

$x^{2} + {(y - - 2)}^{2} = 4$ $x^2 + (y - -2)^2 = 4$

Insert a -0 in the x term:

${(x - 0)}^{2} + {(y - - 2)}^{2} = 4$ $(x - 0)^2 + (y - -2)^2 = 4$

Write the constant as a square:

${(x - 0)}^{2} + {(y - - 2)}^{2} = 2^{2}$ $(x - 0)^2 + (y - -2)^2 = 2^2$

It is easily observed that the center is $(0, - 2)$ $(0, -2)$ and the radius is 2.

How do you find the center and radius of the circle given x2+(y+2)2=4x^2+(y+2)^2=4?