Question 938727
<pre>
We have the center of the circle (h,k) = (4,1), and the
equation of a circle:

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

becomes

{{{(x-4)^2+(y-1)^2=r^2}}}

So all we are lacking is r, the radius.

We sketch the graph approximately:

{{{drawing(400,3600/11,-3,8,-4,5,graph(400,3600/11,-3,8,-4,5),
green(line(2.32,-1.24,4,1)),
line(-14,11,14,-10), circle(4,1,2.8),
circle(4,1,0.15),circle(4,1,0.13),circle(4,1,0.11),circle(4,1,0.09),circle(4,1,0.07),circle(4,1,0.05),circle(4,1,0.03),circle(4,1,0.01),
locate(4,1,"(4,1)"),
locate(4,-2,3x+4y-2=0) 



   )}}}

The green line is a radius, and it is the perpendicular 
distance from the center to the line.  We use this formula:

The perpendicular distance from the point (x<sub>1</sub>,y<sub>1</sub>)
to the line Ax+By+C=0 is given by this formula:

d = {{{abs(Ax[1]+By[1]+C)/sqrt(A^2+B^2)}}}

(x<sub>1</sub>,y<sub>1</sub>) = (4,1)

3x+4y-2=0 is Ax+By+C=0, so A=3, B=4, C=-2

r = {{{abs(3*4+4*1-2)/sqrt(3^2+4^2)}}}

r = {{{abs(12+4-2)/sqrt(9+16)}}}

r = {{{abs(14)/sqrt(25)}}}

r = {{{14/5}}}

So the equation of the circle is:

{{{(x-4)^2+(y-1)^2=(14/5)^2}}}

Edwin</pre>