There's a circle with the equation:

\(\displaystyle x^2 + y^2-6x + 2y -15 = 0\)

Another Circle has the Center (11,14) and radius 8. A Point Q lies on Circle 1 and a Point R lies on Circle 2. Find the shortest possible distance.

My solution was to find the distance between the 2 circles and from there I determined that they intersected. So from there I came to the conclusion that the shorest distance was 0 and this siuation occurs when Q=R.

I assume you need to find the shortest possible distance between the Q and R? And I don't have any idea what methods you used to find the distance between the 2 circles, but these two don't intersect.

The equation for the first circle should be converted to a useful form by completing the square:

\(\displaystyle x^2-6x + y^2+2y = 15\)

\(\displaystyle (x-3)^2-9 + (y+1)^2-1 = 15\)

\(\displaystyle (x-3)^2+(y+1)^2=25\)

So this circle is centered at (3,-1) and has a radius of 5.

Draw a line segment from the center of one circle to the center of another. The length of this segment is:

\(\displaystyle d=\sqrt{(11-3)^2+(14+1)^2}\implies d=17\)

The shortest distance between the circles is this distance minus the two radii, which is 4 units. So the minimum distance between P and Q is 4.