Write an equation of the circle described.
1. Center at the origin; radius 9.

3. Find the center and the radius of the circle (x+2)^+(y-3)^=36.

Thank you ....

2. Hello,

There is a direct way for it, by knowing the equation of a circle. Since I don't remember it, I'll show you how to retrieve it

Any point on the circle will be at distance R from the center, where R is the radius.
Let this point have coordinates (x,y) (yes, it will be the variables ).

Sketch a circle (see the attached picture).
Then let A be $\displaystyle (x_0,y_0)$, that is to say coordinates of the center. Let B the point with coordinates (x,y).
a and b will be two perpendicular lines, intersecting in D.

What you want is that AB is always equal to R, the radius of the circle.

Now, how are we going to get it ? Thanks to the Pythagorean theorem.

The distance AD will be the difference between the absciss of A and B. This means that it will be $\displaystyle AD=(x-x_0)$

The distance BD will be the difference between the ordinates of A and B. So it's $\displaystyle BD=(y-y_0)$

Applying the theorem, we have :

$\displaystyle AB^2=BD^2+AD^2$

Hence $\displaystyle \boxed{R^2=(x-x_0)^2+(y-y_0)^2}$

Take for example your first circle.
$\displaystyle R=9$ and $\displaystyle x_0=0$ and $\displaystyle y_0=0$

So just replace it in the equation

3. Originally Posted by sarayork

Write an equation of the circle described.
1. Center at the origin; radius 9.

You can do (2) similarly with $\displaystyle x_0 = -1$, $\displaystyle y_0 = 2$ and $\displaystyle R = 5$.
For (3) compare Moo's formula for a circle, and obtain the values of $\displaystyle x_0$, $\displaystyle y_0$ and R