Show that $r=\langle x,y\rangle$ satisfies $(r-a)\cdot (r-b)$ iff $(x,y)$ is on a circle

Question

Given that $a=\langle a_1,a_2\rangle$, $b=\langle b_1,b_2\rangle$. How can we show that $r=\langle x,y\rangle$ satisfies $(r-a)\cdot (r-b)=0$ iff $(x,y)$ is on a circle? What would be the center and radius of such a circle in terms of vector operations?

My thoughts:

I really don't know where to start with this question. Can someone help me?

Answer 1

The statement of the problem is ambiguous. It should be:

"Show that for any two vectors $a, b \in \Bbb R ^2$ there exist another vector $p \in \Bbb R ^2$ and a number $R>0$ such that: $(r-a) \cdot (r-b) = 0 \Leftrightarrow \| r-p \| ^2 = R^2$."

Before jumping to the solution, let us work a little bit on the algebraic expression above:

$$(r-a) \cdot (r-b) = \| r \| ^2 - (a+b) \cdot r + a \cdot b = \| r \| ^2 - \frac 1 2 (a+b) \cdot r - \frac 1 2 r \cdot (a+b) + \frac 1 4 \| a+b \| ^2 - \frac 1 4 \| a+b \| ^2 + a \cdot b = \| r - \frac 1 2 (a+b) \| ^2 - \frac 1 4 (\| a \|^2 + 2 a \cdot b + \| b \|^2 - 4 a \cdot b) = \| r - \frac 1 2 (a+b) \| ^2 - \frac 1 4 \| a-b \|^2 .$$

Now, things are clear: just choose $p = \frac 1 2 (a+b)$ and $R = \frac {\| a-b \|} 2$ and you are done.

Answer 2

The dot product of two nonzero vectors is $0$ if and only if they meet at a right angle.

I think it was (?) Thales of Miletus (the founder of philosophy, by some accounts, I think?) who showed that if $a$ and $b$ are two points on a circle that are antipodal to each other then for any point $x$ on the circle, the segment from $a$ to $x$ is perpendicular to the segment from $b$ to $x$.

So see if that leads you somewhere.

Postscript in response to a comment:

The above should suggest that the center of the circle is the midpoint between $a$ and $b$. The midpoint is the average, i.e. $(a+b)/2$. The radius is half the diameter, and the diameter is the distance from $a$ to $b$, and the distance from $a$ to $b$ is $\sqrt{(a-b)\cdot(a-b)}$.

An equation of the circle centered at $(a+b)/2$ with radius $\frac 1 2 \sqrt{(a-b)\cdot(a-b)}$ is $$ \left(r - \frac{a+b}2\right)\cdot\left(r - \frac{a+b}2\right) = \frac 1 4 (a-b)\cdot(a-b). \tag 1 $$ So the question is whether $(1)$ is true if and only if $(r-a)\cdot(r-b)=0$.

Answer 3

Let $A$ be the image of $a$ and $B$ the image of $b$ and $R$ the image of $r$. $r-a$ represents $\vec{AR}$ and $r-b$ represents the vector $\vec{BR}$. $\vec{AR} \cdot \vec{BR}=0 \Leftrightarrow (AR) \perp(BR) \Leftrightarrow$ ARB is a right ange triangle at $R$, $\Leftrightarrow R$ belongs to the circle of diameter $AB$. Edited later :the center is the midpoint of the segment $[AB]$ and has coordinates $(\frac{a_{1}+b_{1}}{2},\frac{a_{2}+b_{2}}{2}))$ and the radius is $\vert AB \vert =\frac{\sqrt{(a_{1}-b_{1})^{2}+(a_{2}-b_{2})^{2}}}{2}$