# Circles

• January 27th 2008, 04:58 PM
heathrowjohnny
Circles
Let $\bold{r}$ be a point on a circle with diameter from $- \bold{a}$ to $\bold{a}$ (so that $|\bold{r}| = |\bold{a}|$). Draw the chord through $\bold{r}$ perpendicular to $\bold{a}$ (the length and direction of this chord is $2 \bold{r}_{a \perp}$). Let $\bold{t} = \bold{r}_{a}$ be the point of intersection of this chord with the diameter. Prove that the square of half the length of the chord is the product of the distance from $-\bold{a}$ to $\bold{t}$ times the distance from $\bold{t}$ to $\bold{a}$: $|\bold{r}_{a \perp}|^{2} = (\bold{t} + \bold{a}) \cdot (\bold{a} - \bold{t})$.
• January 27th 2008, 10:22 PM
heathrowjohnny
So I got $\hat{\bold{a}} = \frac{\bold{a}}{|\bold{a}|}$

and $\bold{t} = \bold{r} \cdot \hat{\bold{a}}$

and $\bold{r}_{a \perp} = \bold{t} - \bold{r}$.

Then $\bold{a} + \bold{r}$ gives one hypotenuse. And so $|\bold{a} + \bold{r}|^{2} = |\bold{r}_{a}|^{2} + |\bold{a}+\bold{t}|^{2}$.

So $|\bold{r} - \bold{a}|^{2} = |\bold{r}_{a}|^{2} + |\bold{a}-\bold{t}|^{2}$.
• January 27th 2008, 11:23 PM
earboth
Quote:

Originally Posted by heathrowjohnny
Let $\bold{r}$ be a point on a circle with diameter from $- \bold{a}$ to $\bold{a}$ (so that $|\bold{r}| = |\bold{a}|$). Draw the chord through $\bold{r}$ perpendicular to $\bold{a}$ (the length and direction of this chord is $2 \bold{r}_{a \perp}$). Let $\bold{t} = \bold{r}_{a}$ be the point of intersection of this chord with the diameter. Prove that the square of half the length of the chord is the product of the distance from $-\bold{a}$ to $\bold{t}$ times the distance from $\bold{t}$ to $\bold{a}$: $|\bold{r}_{a \perp}|^{2} = (\bold{t} + \bold{a}) \cdot (\bold{a} - \bold{t})$.

Hello,

you are dealing with a right triangle (see attachment)

If the legs of a right triangle are a and b and the hypotenuse is c then you know:

$a^2+b^2 = c^2$ ....... and $c = p+q$. That means $c^2 = p^2+2pq+q^2$

Therefore:

$a^2+b^2 = p^2+2pq+q^2$

According to the drawing you have:

$a^2 = h^2 + p^2$ ....... and $b^2 = h^2 +q^2$

So finally you get:

$h^2 + p^2 + h^2 +q^2 = p^2+2pq+q^2~\iff~ 2h^2 = 2pq~\iff~\boxed{h^2 = pq}$