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? 
 A: 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!  
A: Your approach was correct and you are indeed very close to your answer.
From the radical axis equation it is pretty obvious that the centre of the orthogonal circle will lie have coords (2,y).
Because the radius of the orthogonal circle will be the length of tangent to any one of the circle, hence , by finding the length of tangent we can arrive at our answer.
Using Length of Tangent(L) = √S1 , where S1 = the value we get after plugging in the coords of an external point
therefore L = √(y^2 + 15)
now to get get minimum length, you can either differentiate it wrt y or just plug in y=0.
hence minimum radius is  √15
