Question 990689
Equation of circle is
{{{ x^2 + y^2 + ax + 2ay = 3}}}
=> {{{(x+a/2)^2 -a^2/4 + (y+a)^2 -a^2 =3}}}
=> {{{(x+a/2)^2 + (y+a)^2 = 3+5a^2/4}}}
So the center of circle is
(-a/2, -a) and radius is {{{sqrt(3+5a^2/4)}}}

Slope of radius at (0,b) = {{{(-a-b)/(-a/2 -0)}}}
=2(a+b)/a
 SO slope of tangent at (0,b) = (-a/(2(a+b))) (Since radius and tangent are perpendicular and so product of their slopes will be -1)

So let equation of tangent be 
{{{ y = (-a/(2(a+b)))x + c}}}
Since it passes through (0,b)
So,
b=(-a/(2(a+b)))*0 + c
=> b=c
 So the equation of tangent becomes,
{{{ y = (-a/(2(a+b)))x + b}}} --------------Required equation of tangent
 If this tangent has a gradient of 1/4
then, 
{{{(-a/(2(a+b))) = 1/4}}}
=> {{{-2a=a+b}}}
=> {{{3a+b=0}}}  ----(i) ----------------Required relation between a and b

Also, since (0,b) lies on circle,So
0^2 + b^2 + a*0 + 2ab =3
=> b^2 + 2ab - 3 = 0
Putting value of b from eqn. (i) above,
9a^2 -6a^2 - 3 =0
=> a^2 =1
=> a=1 or a = -1
Putting these values of a in original equations of circle, we get
{{{x^2 + y^2 + x + 2y = 3}}}
and
{{{x^2 + y^2 - x - 2y = 3}}}
 as two equations of possible circles.