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What is the shortest distance between two circles, the first having center (5,3) and radius 12 and the other with center (2,-1) and radius 6.

When I draw the circle, I can figure out that the shortest distance between the circles can be calculating by drawing a line through the center point of both of circles, but that is only visual understanding. Is there any theorem behind this to make sure that my visual understanding is correct?

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Hint: Name center of circles as $o(a_1,b_1),o'(a_2,b_2)$ usually take $oo'=d=\sqrt{(a_1-a_2)^2+(b_1-b_2)^2}$ then you have two cirlcle with radius $R,R'$ according to your figure ,minimum distance between two circles is $R-(d+R')$ enter image description here

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    Thank you. My question was how do I ensure that the line which is going to o and o' is the correct line which gives the shortest distance, why can't it be a different line which does not pass through the circle centers?(visually I believe that the line will give the shortest distance, bu t not sure mathematically) – 2017-02-03
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    This concept has a mathematical proof . – 2017-02-03
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    thank you. Can you please provide some more details regarding the mathematical proof? – 2017-02-03
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    I am poor in English ...I Think I can't tell about the proof well ,But i try. when you fix a point in big circle which point in other circle has minimum distance ? then fix that point , and fin minimum distance to big circle . – 2017-02-03
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    See this :https://www.geogebra.org/m/nGxcAbAy – 2017-02-03
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(The following is an adaptation of this related answer to the case of interior circles).

Let $A, B$ be the centers of the large and small circles, respectively. Let $C,D$ be the intersections of $AB$ with the small and large circles, respectively.

Suppose points $C',D'$ existed on the small and large circles, respectively, so that $C'D' \lt CD$. Since $BC=BC'$ and $AD=AD'$ that would imply:

$$ AB+BC'+C'D' \lt AB + BC'+CD = AB + BC + CD = AD = AD' $$

But the shortest distance between two points is the straight line, so the length of the broken line $ABC'D'$ can be no smaller than that of segment $AD'\,$, with equality iff $C'=D$, $D'=D$.