Question 704675
Find an equation of the circle with center at the origin and passing through (-3,-4) in the form of (x-A)² + (y-B)² = C where A, B, C are constants 
<pre>
{{{drawing(400,400,-6,6,-6,6,graph(400,400,-6,6,-6,6),locate(.2,.4,"(0,0)"),

circle(0,0,5),circle(-3,-4,.1),locate(-4.9,-3.7,"(-3,-4)") )}}}

It has center (A,B) = (0,0) so we can substitute A=0 and B=0 into:

  (x-A)² + (y-B)² = C

  (x-0)² + (y-0)² = C

And since it goes through (-3,-4)

we can substitute x=-3, and y=-4 

(-3-0)² + (-4-0)² = C

(-3)² + (-4)² = C

       9 + 16 = C

           25 = C

Now we can substitute 25 for C:

  (x-0)² + (y-0)² = C

  (x-0)² + (y-0)² = 25

That's the answer showing the 0's for A and B.
You can erase the 0's and just have

          x² + y² = 25

Unless your teacher want's you to leave the 0's 
to show what A and B are.

Edwin</pre>

Edwin</pre>