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EXTRA

DONE Unit 40: Center Radius Form of a Circle

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.

$$\boldsymbol(\boldsymbol x\boldsymbol-\boldsymbol h\boldsymbol)^{\mathbf2}\boldsymbol+\boldsymbol(\boldsymbol y\boldsymbol-\boldsymbol k\boldsymbol)^{\mathbf2}\boldsymbol=\boldsymbol r^{\mathbf2}$$

(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!).