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

1 Answer
Patrick H.
Jun 5, 2018

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(x−h)2+(y−k)2=r2

where hh is the x-value of the origin, kk is the y-value of the origin, and rr 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 hh and kk have. Therefore, our origin is (2,-3)(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
(x-2)^2+(y+3)^2=4(x−2)2+(y+3)2=4

For our radius, we take the square root of r^2r2 value:

r^2=4r2=4
r=2r=2
Note that -2−2 is not an answer because it is not possible to have a negative distance in a graph.

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