Question 258586
The ordered pair (-3,2) lies on a circle with center (0,3). What is the equation of the circle? i dont get how to work this problem out can i please get help?
<pre><font size = 4 color = "indigo"><b>
Momorize these two facts:

1.  The equation of the circle with center (h,k) and radius, r, is:

(x-h)<sup>2</sup> + (y-k)<sup>2</sup> = r<sup>2</sup>

2. The distance, d, between the points (x<sub>1</sub>,y<sub>1</sub>) and
   (x<sub>2</sub>,y<sub>2</sub>) is given by this formula:

{{{d = sqrt( (x[2]-x[1])^2 + (y[2]-y[1])^2 ) }}}  

Plot those points:

{{{drawing(200,200,-5,5,-2.5,7.5, graph(200,200,-5,5,-2.5,7.5),

line(-3+.1,2,-3-.1,2), line(-3,2+.1,-3,2-.1), line(-3+.1,2+.1,-3-.1,2-.1), line(-3+.1,2-.1,-3-.1,2+.1), locate(-3,2,"(-3,2)"), locate(0,3,"(0,3)"),

line(0+.1,3,0-.1,3), line(0,3+.1,0,3-.1), line(0+.1,3+.1,0-.1,3-.1), line(0+.1,3-.1,0-.1,3+.1) )}}}

Now if we connect them that will be a radius, and we can draw the circle
by putting the point of a compass on (0,3) and the pencil of the compass
on (-3,2) and swinging it around.

{{{drawing(200,200,-5,5,-2.5,7.5, graph(200,200,-5,5,-2.5,7.5),

line(-3+.1,2,-3-.1,2), line(-3,2+.1,-3,2-.1), line(-3+.1,2+.1,-3-.1,2-.1), line(-3+.1,2-.1,-3-.1,2+.1), locate(-3,2,"(-3,2)"), locate(0,3,"(0,3)"),
red(line(0,3,-3,2)), circle(0,3,sqrt(10)),
line(0+.1,3,0-.1,3), line(0,3+.1,0,3-.1), line(0+.1,3+.1,0-.1,3-.1), line(0+.1,3-.1,0-.1,3+.1) )}}}

If we find the distance between those two points we will have the radius r.

So we use this formula to find that distance:

{{{d = sqrt( (x[2]-x[1])^2 + (y[2]-y[1])^2 ) }}} 

where (x<sub>1</sub>,y<sub>1</sub>) = (0,3) and (x<sub>2</sub>,y<sub>2</sub>) = (-3,2)

Substituting:

{{{d = sqrt( (-3-0)^2 + (2-3)^2 ) }}}

{{{d = sqrt( (-3)^2 + (-1)^2 ) }}} 

{{{d = sqrt( 9 + 1 ) }}}

{{{d = sqrt( 10 ) }}}

So d is the radius r.

Now we are ready to substitute into

{{{(x-h)^2+(y-k)^2=r^3}}} with center {{{"(h,k)"="(0,3)"}}} and  {{{r=sqrt(10)}}}.

{{{(x-0)^2+(y-3)^2=(sqrt(10))^2}}}

{{{x^2+(y-3)^2=10}}}

Edwin</pre>