Question 697785
<pre>
{{{drawing(800,10000/17,-.5,3,-.5,2,graph(800,10000/17,-.5,3,-.5,2),line(8,4,-7,-3.5),locate(2.3,1.4,r),
circle(12/5,6/5,1/sqrt(5)), green(line(0,0,2.136130674,1.561071986),
line(0,0,2.530535993,.7722613478)),red(line(12/5,6/5,2.136130674,1.561071986),
line(12/5,6/5,2.530535993,.7722613478)),locate(1.5,.45,sqrt(7)),
locate(1.2,1,sqrt(7)), locate(2.45,1.2,"C(h,k)"),locate(2.5,1.05,r),
locate(.05,-.05,"O(0,0)"),locate(2.53,.77,T),locate(2.05,1,"(2,1)")

 )}}}

Let the equation be 

(1)     (x-h)² + (y-k)² = r²

and since it is on the line 2y=x,

(2)     2k = h

(2,1) is on the circle and also on the line 2y=x

So (2,1) satisfies the equation of the circle, so

        (2-h)² + (1-k)² = r²
        4-4h+h²+1-2k+k² = r²
(3)         h²+k²-4h-2k = r²-5 

By the distance formula OC² = h²+k² 

OCT is a right triangle because a tangent is perpendicular
to the radius drawn to the point of tangency. So using
the Pythagorean theorem:

        OC² = TC² + OT² 
        h²+k² = r²+{{{(sqrt(7))^2}}} 

(4)       h²+k² = r²+7     

Subtracting equation (4) from equation (3) gives

         -4h-2k = -12  or

(5)       2h+k = 6

Using (2), (5) becomes

       2(2k)+k = 6
          4k+k = 6
            5k = 6
             k = {{{6/5}}}

Using (2), h = 2·{{{6/5}}}
           h = {{{12/5}}}

So the center C(h,k) = C({{{12/5}}},{{{6/5}}})

We just need to find r²

Using (4)

          h²+k² = r²+7
         {{{(12/5)^2}}}+{{{(6/5)^2}}} = r²+7
         {{{144/25}}}+{{{36/25}}} = r²+7
         {{{180/25}}} = r²+7
         {{{36/5}}} = r²+7
         {{{36/5}}}-7 = r²
         {{{36/5}}}-{{{35/5}}} = r²
         {{{1/5}}} = r²

So we can write the equation of the circle as

        (x-h)² + (y-k)² = r²
        {x-{{{12/5}}})² + (y-{{{6/5}}})² = {{{1/5}}}

If you like you can square those out, collect terms and clear of 
fractions and you'll end up with

        5x² + 5y² - 24x - 12y + 35 = 0

Edwin</pre>