Question 1168435
If the circle is tangent to the line -3x+2y+1=0 at the point (1, 1), and the
center is an the line x+y-1=0. Find the general equation of the circle.
<pre>
{{{drawing(400,400,-10,10,-10,10,

graph(400,400,-10,10,-10,10,(3x-1)/2),circle(-2,3,sqrt(13)),
circle(-2,3,.2),circle(1,1,.2),green(line(-2,3,1,1)),
graph(400,400,-10,10,-10,10,-x+1) )}}}

Let the equation of the circle be:

{{{(x-h)^2+(y-k)^2=r^2}}}

Then the tangent point (1,1) is on the circle, so

{{{(1-h)^2+(1-k)^2=r^2}}}

The center (h,k) lies on the line 

{{{x+y-1=0}}}, so

{{{h+k-1=0}}}

The perpendicular distance from the center (h,k), to the tangent line, 
which is
{{{-3x+2y+1=0}}}

is the radius r (in green), so

{{{(-3h+2k+1)/sqrt(13)="" +- r}}}

So we have the system of three equations in three unknowns:

{{{system((1-h)^2+(1-k)^2=r^2,h+k-1=0,(-3h+2k+1)/sqrt(13)="" +- r)}}}

Can you find the solution?

The solution is (h,k,r) = (-2,3,√13)

So the equation of the circle is

(x+2)<sup>2</sup> + (y-3)<sup>2</sup> = 13 

If you have trouble finding the solution to the system,
tell me about it in the thank you message below, and I'll
get back to you by email.  No charge.

Edwin</pre>