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It seems obvoius that in a circle the shortest distance between any point inside the circle and it's margin(circumference) is found on the line that connects the center of the circle and the point inside the circle.. but how can i prove this?

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    $\begingroup$ Draw a picture. Connect your point to the center, and connect your point $p$. to the circumference twice: once along the radius which passes through $p$, and once to any other point on the circumference. Draw a triangle. Use the triangle inequality. $\endgroup$ – uniquesolution Jun 26 '17 at 12:49
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Hint:

Without loss of generality you can consider a circumference center at the origin and radius $R$ and a point $P$ on the $x-$axis with coordinates $P=(a,0)$ ( with $|a|<R$). A point on the circumference is $Q=(R\cos \theta, R\sin \theta)$.

So you have to minimize the function $\overline{PQ}^2=(a-R\cos \theta)^2+(R\sin \theta)^2$

It is easy!

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Without loss of generality, let the circle be centered at origin and be of radius $r$. Let $P \equiv (x_0,y_0)$ be a point inside the circle and let $Q \equiv (x,y)$ be an arbitrary point on the circle which will satisfy $x^2+y^2=r^2$.

Now, the length of $PQ$ (or its square) has to be minimized with respect to $(x,y)$, that is,

$$\min\limits_{(x,y)}\,\, (x-x_0)^2 + (y-y_0)^2 \, \text{ s.t. } \, x^2+y^2=r^2$$

The Lagrangian will be,

$$L(x,y,\lambda) = (x-x_0)^2 + (y-y_0)^2 + \lambda(x^2+y^2-r^2)$$

$$\frac{\partial L(x,y,\lambda)}{\partial x} = 2(x-x_0) + 2x\lambda = 0 \implies x = \frac{x_0}{1+\lambda}$$

$$\frac{\partial L(x,y,\lambda)}{\partial y} = 2(y-y_0) + 2y\lambda = 0 \implies y = \frac{y_0}{1+\lambda}$$

$$\frac{\partial L(x,y,\lambda)}{\partial \lambda} = x^2+y^2-r^2 = 0$$

From the first two equations, one can conclude that $\frac{y}{x} = \frac{y_0}{x_0}$ which implies that $(x,y)$ lies on the line passing through origin (center) and $(x_0, y_0)$.

Note, one must compute the values of $(x,y)$ and then compute the Hessian as in this answer corresponsing to the computed values to finally conclude that a minima or maxima is achieved.

$$\left(\frac{x_0}{1+\lambda}\right)^2 + \left(\frac{y_0}{1+\lambda}\right)^2 - r^2 = 0 \implies (1+\lambda)^2 = \frac{x_0^2+y_0^2}{r^2} \implies 1+\lambda = \pm\frac{\sqrt{x_0^2+y_0^2}}{r}$$

One of the value of $\lambda$ will give minima and other will give maxima, however, in both cases, the coordinates $(x,y)$ will lie on the line passing through origin and $(x_0, y_0)$.

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Draw a circle with center in your point and radius equal to distance to its "circumference", which you obtain by stated approach. The radius $r$ is now equal to $a+b$, where $a$ is distance from center of the bigger circle to your point and $b$ is the rest. Now any other point on the smaller circle has a distance $b$ from point P, and by triangular inequality, $a+b>c$, so any other point on the smaller circle is closer to the center than any other point on the big circle.

Therefore those two circles are only touching and there is no intersection between them, so it has to be the closest point on the big circle, all points closer lie inside the smaller circle.

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