How do you find the center and radius of #(x-2)^2+(y+3)^2=4#?

1 Answer
Jun 5, 2018

Answer:

In order to find the center and radius, we must first understand how a standard circle equation is written.

Explanation:

The standard form for a circle is written as:

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

where #h# is the x-value of the origin, #k# is the y-value of the origin, and #r# is the radius of the circle. Firstly, let's take out our origin from the equation. A way I like to imagine the origin is to switch the sign that #h# and #k# have. Therefore, our origin is #(2,-3)#. Let's plug in our origin back into the equation to make sure:

#(x-(2))^2+(y-(-3))^2=4#
#(x-2)^2+(y+3)^2=4#

For our radius, we take the square root of #r^2# value:

#r^2=4#
#r=2#
Note that #-2# is not an answer because it is not possible to have a negative distance in a graph.