If a circle lies entirely inside another, but is not concentric with it, can their radical axis be closer to the smaller circle?

Question

We say that if circle $S_1$ lies completely inside circle $S_2$ but are not concentric then the radical axis will surely lie outside both the circles, but can we say that the radical axis will be closer to the inner smaller circle?

Answer 1

As you can see, the radical axis will always be outside the two circles in this case. This is easy to prove. Suppose it is inside, then for sure it will intersect $O_1$ at two points and since the two circles are not intersecting, at that point of intersection, the powers will be $0$ for $O_1$ and non-zero for $O_2$ which contradicts the definition of Radical axis. And for the second question, it will surely be closer to the smaller circle's center because:

Suppose $O_1O_2$ meet radical axis at $X$, then $XO_2^2-r_2^2=XO_1^2-r_1^2\implies X0_2<XO_1$ because, $r_1>r_2$.

Answer 2

The radical axis of circles

$$ (x - H)^2 + y^2 == R^2 $$

$$ (x - h)^2 + y^2 == r^2 $$

is obtained by subtracting one equation of circles from the other as a straight line parallel to x-axis

$$ x= [(H + h) - (R^2 - r^2)/(H - h)]/2$$

The radical axis lies either right or left of radical axis intersection with $y=0$ occur in the order depending on $$ h>H \, or \, h<H $$

( axis, big circle ,small circle )

or

( small circle , big circle ,axis )

i.e., bigger circle always shifts towards radical axis.