The radical center of the three circles is at the origin. The equation of the two of the circles are  $x^2+y^2=1and{x}^2+y^2+4x+4y-1=0$. If the third circle passes through the points $(1,1)$  and $(-2,1)$  : and its radius can be expressed in the form of $\frac{p}{q}$ , where p and q are relatively prime positive integers. Find the value of  $(p+q)$.

By using system of circles any circle passing through (1,1) and (-2,1) is 
$(x-1)(x+2)+{(y-1)}^2+\lambda (y-1)=0$
Given circles    $x^2+y^2-1=0$     
Now radical axis of (1) and (2) is 
$(x-2y)+\lambda (y-1)=0$
$\because$ Radical center of given circles is(0,0). 
So, eq. (3) is passing through(0,0).
$\therefore \lambda =0$  
Put  $\lambda =0$ in eq. (1) we get required circle.
Radius of the required circle is $\frac{3}{2}$