# Prove this property for an arbitrary circle

Prove that in an arbitrary circle, the point on the circle closest to the origin must lie on the extended line connecting the circle's centre and the origin.

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Think about the relation between that line and the tangent to the circle at that point. –  Gerry Myerson Nov 6 '12 at 11:55
It can't be a completely arbitrary circle if there's a "the" line connecting the circle's center and the origin. –  Henning Makholm Nov 6 '12 at 12:02
Sorry for the confusion. I meant the extended line connecting the circle's center and the origin. –  hollow7 Nov 6 '12 at 12:09
I don't know what an extended line is. I think what you're calling an "extended line" is usually called a line, and what you may be thinking of as an "unextended line" is usually called a line segment. I think Henning's point was that if there is a single line through the centre and the origin, as indicated by the definite article, then the centre can't be the origin, and hence the circle isn't completely arbitrary. –  joriki Nov 6 '12 at 15:14

## 5 Answers

Let the equation of the circle be $(x-a)^2+(y-b)^2=r^2$

So, any point on the circle can be $P(r\cos \theta+a,r\sin \theta+b)$

If the distance$(d)$ from the point $(r\cos \theta+a,r\sin \theta+b)$ from the origin,

$d^2=(r\cos \theta+a)^2+(r\sin \theta+b)^2=r^2+a^2+b^2+2r(a\cos \theta+b\sin \theta)=r^2+a^2+b^2+2r\sqrt{a^2+b^2}\cos(\theta-\arctan\frac b a)$

This will be minimum if $\cos(\theta-\arctan\frac b a)=-1$ $\implies \theta-\arctan\frac b a=(2n+1)\pi$ where $n$ is any integer.

So, $\tan \theta=\tan\{(2n+1)\pi+\arctan\frac b a\}=\frac b a\implies b\cos \theta=a\sin \theta$

Now the area of the triangle with vertices $(0,0),(a,b),(r\cos \theta+a,r\sin \theta+b)$ is $$\frac 1 2\det\begin{pmatrix} 0 & 0 & 1 \\ r\cos \theta+a & r\sin \theta+b & 1 \\ a & b&1\end{pmatrix}=\frac 12 \left(b(r\cos \theta+a)-a(r\sin \theta+b)\right)=\frac{r(b\cos \theta-a\sin \theta)}2=0$$

So, $(0,0),(a,b),(r\cos \theta+a,r\sin \theta+b)$ are co-linear for the minimum distance of $P$ from the origin $O(0,0)$ for any $a,b,r$.

A little generalization:

If the distance of $P(r\cos \theta+a,r\sin \theta+b)$ from $Q(c,d)$ is $D,$

$D^2=(r\cos \theta+a-c)^2+(r\sin \theta+b-d)^2$ $=r^2+(a-c)^2+(b-d)^2+2r\{(a-c)\cos \theta+(b-d)\sin \theta\}$

Applying the same method,the distance will be minimum is $(b-d)\cos \theta=(a-c)\sin \theta$

Now the area of the triangle with vertices $(c,d),(a,b),(r\cos \theta+a,r\sin \theta+b)$ is $$\frac 1 2\det\begin{pmatrix} c & d & 1 \\ r\cos \theta+a & r\sin \theta+b & 1 \\ a & b&1\end{pmatrix}$$ $$=\frac 12\det\begin{pmatrix} c-a & d-b & 1-1 \\ r\cos \theta+a-a & r\sin \theta+b-b & 1-1 \\ a & b&1\end{pmatrix}=\frac{r\{(a-c)\sin\theta-(d-b)\cos\theta\}}2=0$$

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Case 1: $O$ outside.

Let $P$ be any point on the circle, and $C$ the centre. By the triangle inequality

$$PO \geq CO-PC= CO-R$$

with equality if and only if $CPO$ colinear in this order.

Thus the minimum of $PO$ is exactly $CO-R$, and is attained exactly when $P$ is the intersection of $CO$ with the circle.

Case 2: $O$ on the circle. The problem is obvious.

Case 3: $O$ inside but $o \neq C$. Exactly as in the first case, by the triangle inequality

$$PO \geq PC-CO=R- CO \,,$$ and the rest is identical with the first case.

Case 4 $O=C$, the problem doesn't make sense.

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1.- If the given circle's center is the origin then there's nothing to prove: all the circle's points are equidistant to the origin;

2.- Otherwise, draw a circle (let's call it TC = tangent circle) with center at the origin and tangent, internally or externally, to the given one (let's call this one OC=original circle).

Pay attention to the fact that in this case the minimal distance from the OC to the origin is precisely the TC's radius (a diagram will help you a lot to check this trivial fact), and since the only point on the OC whose distance to the origin equals this radius is the tangency point of both circle, we're done!

Of course, the above is just a re-wording of the well known theorem in Euclidean geometry:

Theorem: If two circles are tangent to each other, then the straight line joining their centers passes through their tangency point.

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Let the origin be $O$ and the circle's center be $C$.

• For $O$ on the circumference, we're done. For $O=C$, the statement as given is vacuous (as @Henning notes) because there's no such thing as "the [unique] line connecting [$C$] and [$O$]".

• For $O\ne C$ in the circle's interior, let $P$ be the point where ray $CO$ meets the circle's circumference, and let $Q\ne P$ be any other point on the circumference. $\triangle CPQ$ is isosceles with acute angles at $P$ and $Q$; as point $O$ lies between $C$ and $P$, ray $QO$ lies between rays $QC$ and $QP$, so that $\angle OQP < \angle CQP \cong \angle OPQ$. By what I like to call the "Three Bears Theorem"[*] applied to $\triangle OPQ$, we must have $|OP| < |OQ|$.

• For $O$ in the circle's exterior to the circle, with $P$ and $Q$ as described above, $\triangle CPQ$ is again isosceles with, necessarily, an acute angle at $P$; therefore, $\triangle OPQ$ has an obtuse angle at $P$, so that the Three Bears again give us $|OQ|>|OP|$.

[*] The "Three Bears Theorem": The Poppa Bear's angle is opposite the Poppa Bear's side; the Momma Bear's angle is opposite the Momma Bear's side; and the Baby Bear's angle is opposite the Baby Bear's side.

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I would like to approach this as an optimization problem.

The equation of the circle with radius r is $(x-a)^2+(y-b)^2 = r^2 \ge 0$. For any point on the circle $(x_1, y_1)$ such that $(x_1-a)^2+(y_1-b)^2 = r^2$ the distance between this point and the origin is $d$ such that $d^2 = x_1^2+y_1^2$. The task is to prove that the optimization problem

$$\min_{X\in\mathbb R} \left\{\begin{array}{ll} f(x_1,y_1)=x_1^2+y_1^2\\ \text{s.t.}\\ g(x_1, y_1)=(x_1-a)^2+(y_1-b)^2 = r^2\\ \end{array} \right.$$ yield a solution $(x', y')$ that falls on the line joining the centre and the origin; $ay=bx$ that is it must satisfy $$ay'=bx'$$

The function $f(x_1,y_1)$ is convex and the constraint function $g(x_1, y_1)$ is also convex, so there exist a global minimum to the problem. Using the Lagrange multiplier,

$$\Lambda(x_1,y_1,\lambda) = f(x_1,y_1) + \lambda \cdot \Big(g(x_1,y_1)-r^2\Big)$$ $$\Lambda(x_1,y_1,\lambda) = x_1^2+y_1^2+\lambda \Big((x_1-a)^2+(y_1-b)^2 -r^2 \Big)$$ $$\Lambda(x_1,y_1,\lambda) =x_1^2+\lambda x_1^2+y_1^2+\lambda y_1^2-2 a \lambda x_1-2 b \lambda y_1+(a^2+b^2-r^2)\cdot\lambda$$ $$\nabla_{x_1, y_1, \lambda}\Lambda = 0$$ will give us a series of equation and we solve for $x_1$, $y_1$ and $\lambda$ to get

$$\begin {array}{ll} x_1 = a\lambda /(\lambda+1)\\ y_1 = b\lambda /(\lambda+1)\\ (a-x_1)^2+(b-y_1)^2-r^2 =0 \\ \end{array}$$

the last one gives $$\lambda=\pm \sqrt{\cfrac{a^2+b^2}{r^2}}-1$$ with equivalent values for $x_1$ and $y_1$ $$x_1=a\left (1\pm\sqrt{\cfrac{r^2}{a^2+b^2}} \ \right)$$ $$y_1= b\left (1\pm\sqrt{\cfrac{r^2}{a^2+b^2}} \ \right)$$

Depending on the triplet $(a, \ b, \ r)$ we have both the maximum and the minimum points.

For the couples $(x_1, y_1)$ it is obvious that $$ay_1 = a\cdot b\left (1\pm\sqrt{\cfrac{r^2}{a^2+b^2}} \ \right)=b\cdot a\left (1\pm\sqrt{\cfrac{r^2}{a^2+b^2}} \ \right) = bx_1$$ which satisfy the condition above.

The line not only contain the point that gives the minimum distance but also the point that gives the maximum distance to the origin.

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