Find the minimum radius of the circle which is orthogonal to two given circles Problem: 
Find the minimum radius of the circle which is orthogonal to both the circles $x^2+y^2-12x+35=0$ and $x^2+y^2+4x+3=0$ . 
Solution:
Let the equations : 
$x^2+y^2-12x+35=0\tag i$ 
and   
$x^2+y^2+4x+3=0\tag {ii}$ 
Equation of radical axis of $(i)$ and $(ii)$ is $-16x +32=0 \Rightarrow x =2$ 
It intersects the line joining the center, i.e., $y =0$ at the point $(2,0)$. 
Question: 
How do I find the minimal radius of the circle? 

Orthogonal circles are defined as circles that cut one another at $90^{\circ}$.
Note: The product of the gradients of two perpendicular lines is equal to $-1$ and that the gradient of the tangent to a curve at a point is equal to the gradient of the curve at that point.  
A.T.Q.
As you have correctly concluded, the equation of the required circle is of the form $(x-2)^2+y^2=a^2$ or $y=\pm \sqrt {a^2-(x-2)^2}$. Instead of worrying about both given circles, we will consider one (think about reflection in $x=2$) which I take to be the second one, which simplifies to be: $y=\pm \sqrt {1-(x+2)^2}$.  
$$\frac d{dx} \pm \sqrt {a^2-(x-2)^2} = \mp \frac {x-2}{\sqrt {a^2-(x-2)^2}}$$  
$$\frac d{dx} \pm \sqrt {1-(x+2)^2} = \mp \frac {x+2}{\sqrt {1-(x+2)^2}}$$  
As $m_1*m_2=-1$,
$$\mp \frac {x-2}{\sqrt {a^2-(x-2)^2}}*\mp \frac {x+2}{\sqrt {1-(x+2)^2}}=-1$$
$$x^2-4=-\sqrt {a^2-(x-2)^2}*\sqrt {1-(x+2)^2}$$
$$-(x^2-4)=\sqrt {a^2-a^2(x+2)^2-(x-2)^2+(x^2-4)^2} $$
$$(x^2-4)^2= a^2-a^2(x+2)^2-(x-2)^2+(x^2-4)^2$$
$$a^2-a^2(x^2+4x+4)=x^2-4x+4$$
$$a^2(-x^2-4x-3)=x^2-4x+4$$
$$a^2=-\frac {x^2-4x+4}{x^2+4x+3}$$   
The minimum value of $a^2$ is $15$ and so the minimum radius of the circle which is orthogonal to both the given circles is $\sqrt 15 \approx 3.872983346$. Hope this very late answer helps!  
