If you don’t know it already, this is another formula you will have to memorize. It describes every point on a circle, as well as the center and radius of the circle.

$$boldsymbol(boldsymbol xboldsymbol-boldsymbol hboldsymbol)^{mathbf2}boldsymbol+boldsymbol(boldsymbol yboldsymbol-boldsymbol kboldsymbol)^{mathbf2}boldsymbol=boldsymbol r^{mathbf2}$$

(*h,k*) describes the *x* and* y* coordinates of the center of the circle, and *r* is simply the radius. *x* and *y* in the equation represent any point going around the circle.

If you were given the equation of a circle that looked like this:

$$x^2+y^2=144$$

You would know the center of the circle is at (0,0), and it has a radius of 12. That equation is the same as:

$$(x-0)^2+(y-0)^2=144$$

Our *h* and *k* values are thus 0, and the radius is (sqrt{144}) or 12.

**Strategy****:**

**Know the formula and understand all components of it.**

Another example that is slightly more difficult:

$$(x-2)^2+(y+3)^2=50$$

Here, our center would be at (2,-3) and the radius would be (sqrt{50}). Note the *y* value is negative because it is (y+3) in the equation, where our standard form says (y-k).

If you are given something that looks close to, but not quite standard form, you need to manipulate it until you get it looking just like standard form.

$$2(x-2)^2+2(y+3)^2-20=50$$

We would first add 20 to both sides, then divide by 2:

$$(x-2)^2+(y+3)^2=35$$

Now you can determine the center and radius for this circle.

What if you are given a graph and asked what the equation of the circle is?

Easy! We just find the center of the circle is (-2, 2) and count out that the radius is 3. Since we have our formula memorized, we just plug in those three values to get

$$(x+2)^2+(y-2)^2=9$$

In the next unit, we’ll cover one more type of circle problem (the hardest kind!).

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