Question 990644
Since centre lies on x=2y,
Let the centre be (2a,a)
Eqn of circle will be
{{{(x-2a)^2 + (y-a)^2 = r^2}}}
Since it passes through (2,5), putting it we get,
{{{(2-2a)^2 + (5-a)^2 = r^2}}}
=> {{{4a^2-8a+4 + a^2-10a+25 = r^2}}}
=> {{{5a^2-18a+29 = r^2}}}----------------------------------------(i)
Now length of tangent from origin is sqrt(7)
So, using pythagoras theorem,
 (Distance of centre from origin)^2 = (length of tangent)^2 + (radius)^2
=> {{{(2a-0)^2+ (a-0)^2 = 7 + r^2}}}
=> {{{5a^2-7 = r^2}}}-----------------------------(ii)
Subtracting (i) from (ii),
{{{18a-36 = 0}}}
=> {{{a=2}}}
Putting back in eqn (ii),
{{{r^2=13}}}
Putting both values in eqn of circle we get,
{{{(x-4)^2 + (y-2)^2 = 13}}}
=> {{{x^2-8x+y^2-4y+7=0}}}

P.S. - "The Elements of Coordinate Geometry" by S. L. Loney