How do I measure the distance between two points on a euclidean graphed circle?

**Egoist** · August 29th 2009, 07:21 PM

So I have a circle defined in a euclidean plane, how do I measure the distance along the circle between any two given points on the circle?

The specific circle that I'm looking at right now is defined by the equation:

(x-1280)^2 + (y-800)^2 = 422500

thank you.

**TKHunny** · August 29th 2009, 07:27 PM

Have you considered the radius, the value of $\pi$ , and the proportion of the circle cut out by the wedge formed by the center and the two points?

**Soroban** · August 29th 2009, 10:05 PM

Hello, Egoist!

So I have a circle defined in a euclidean plane.
How do I measure the distance along the circle
between any two given points on the circle?

The specific circle that I'm looking at right now is defined by the equation:
. . $(x-1280)^2 + (y-800)^2 \:=\: 422500$

The center of the circle is: . $C(1280,800)$
The radius is: . $r \:=\:650$

Suppose the two given points are: . $P(x_1,y_1),\;Q(x_2,y_2)$

Find the slope of $CP\!:\;\;m_1 \:=\:\frac{y_1-800}{x_1-1280}$

Find the slope of $CQ\!:\;\;m_2 \:=\:\frac{y_2-800}{x_2-1280}$

The angle $\theta$ between the two radii is given by: . $\tan\theta \:=\:\frac{m_2-m_1}{1 + m_1m_2}$
. . Find $\theta$ in radians.

The arc length between $P$ and $Q$ is: . $s \;=\;r\theta$

**Egoist** · August 30th 2009, 12:11 PM

Originally Posted by Soroban

Hello, Egoist!

The center of the circle is: . $C(1280,800)$
The radius is: . $r \:=\:650$

Suppose the two given points are: . $P(x_1,y_1),\;Q(x_2,y_2)$

Find the slope of $CP\!:\;\;m_1 \:=\:\frac{y_1-800}{x_1-1280}$

Find the slope of $CQ\!:\;\;m_2 \:=\:\frac{y_2-800}{x_2-1280}$

The angle $\theta$ between the two radii is given by: . $\tan\theta \:=\:\frac{m_2-m_1}{1 + m_1m_2}$
. . Find $\theta$ in radians.

The arc length between $P$ and $Q$ is: . $s \;=\;r\theta$

Thank you very much! BTW, as you can see, I'm new here, how do I "officially" thank you?

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