How do you find the center and radius of the circle given $x^2+(y-sqrt3)^2+4x=25$ ?

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Answer 1 · 2016-11-18T22:53:08

Please see the explanation for steps leading to:

The center is $(-2, sqrt(3))$ and the radius is $sqrt(29)$

The standard form for the equation of a circle is:

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

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

Please notice that (y - sqrt(3))^2 is already in that form and we can see that $k = sqrt(3)$ .

Add $h^2$ to both sides of the given equation:

$x^2 + 4x + h^2 + (y - sqrt(3))^2 = 25 + h^2$

When we expand the x square in the standard form we get:

$(x - h)^2 = x^2 - 2hx + h^2$

We can find the value of h by setting the middle term from the standard form equal to the middle term of our equation:

$-2hx = 4x$

$h = -2$

This means that we can substitute $(x - -2)^2$ for the terms on the left and 4 for $h^2$ on the right:

$(x - -2)^2 + (y - sqrt(3))^2 = 25 + 4$

Combine the right side:

$(x - -2)^2 + (y - sqrt(3))^2 = 29$

Write the 29 as $(sqrt(29))^2$

$(x - -2)^2 + (y - sqrt(3))^2 = (sqrt(29))^2$

In this form we can see the center and radius by observation:

The center is $(-2, sqrt(3))$ and the radius is $sqrt(29)$

How do you find the center and radius of the circle given x^2+(y-sqrt3)^2+4x=25x^2+(y-sqrt3)^2+4x=25?