Question 60757
<pre><font size = 5><b>Find the equation of the circle that passes through the 
origin and has its center at (-3,-4).

The equation of a circle with center (h,k) and radius r is

(x - h)² + (y - k)² = r²

We do not know the radius r, but we do know h and k,

(h, k) = (-3, -4), so we have

(x + 3)² + (y + 4)² = r²

Now since we are told that the circle passes through the
origin, then we can substitute (x, y) = (0, 0)

(x + 3)² + (y + 4)² = r²
(0 + 3)² + (0 + 4)² = r²
            3² + 4² = r²
             9 + 16 = r² 
                 25 = r²
                  5 = r

So the equation is

(x + 3)² + (y + 4)² = 5²

(x + 3)² + (y + 4)² = 25

Your teacher may want you to multiply that out and get

x² + y² + 6x + 8y = 0

{{{ graph( 300, 300, -10, 4, -10, 4, sqrt(25-(x+3)^2)-4, -sqrt(25-(x+3)^2)-4,-4+sqrt(.1-(x+3)^2),-4-sqrt(.1-(x+3)^2) ) }}}

Edwin</pre>