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)#