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 April 12th, 2013, 11:54 AM #1 Newbie   Joined: Apr 2013 Posts: 2 Thanks: 0 Finding a point on a circle. I am trying to get a point on a circle (point C) and all I know is the center point of the circle A (let's say at 0,0), the radius (20), and a point outside of the circle B (let's say at 46,53). How would I go about finding point C?
 April 12th, 2013, 01:29 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Finding a point on a circle. Is C the point on segment AB that is on the circle?
 April 12th, 2013, 01:39 PM #3 Newbie   Joined: Apr 2013 Posts: 2 Thanks: 0 Re: Finding a point on a circle. Yes, that is correct.
April 12th, 2013, 01:50 PM   #4
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Re: Finding a point on a circle.

Hello, nickqqqq1!

Quote:
 I am trying to get a point on a circle (point C) and all I know is the center point of the circle A (let's say at 0,0), the radius (20), and a point outside of the circle B (let's say at 46,53). How would I go about finding point C?

$\text{I will assume that the line }AB\text{ intersects the circle at point C.}$

Code:
                |             B
* * *         o (46,53)
*     |     *   *
*       |       o C
*        |     *  *
|   *
*         | *       *
- - * - - - - o - - - - * - -
*       A |         *
|
*        |        *
$\text{The equation of the circle is: }\:x^2\,+\,y^2 \:=\:20^2\;\;[1]
\text{The equation of }AB\text{ is: }\:y \:=\:\frac{53}{46}x\;\;[2]$

$\text{Substitute [2] into [1]: }\:x^2\,+\,\left(\frac{53}{46}x\right)^2 \:=\:20^2 \;\;\;\Rightarrow\;\;\;x^2\,+\,\frac{2809}{2116}x^ 2 \:=\:400$

$\text{Multiply by 2116: }\:2116x^2\,+\,2809x^2\:=\:846,400 \;\;\;\Rightarrow\;\;\; 4925x^2 \:=\:846,400$

[color=beige]. . . . . . . . . . . . . . [/color]$x^2 \:=\:\frac{846,400}{4925} \;\;\;\Rightarrow\;\;\; x \:=\:\sqrt{\frac{846,400}{4925}} \:=\:\frac{184}{\sqrt{197}}$

$\text{Substitute into [2]: }\:y \:=\:\frac{53}{46}\left(\frac{184}{\sqrt{197}}\rig ht) \:=\:\frac{212}{\sqrt{197}}$

$\text{Therefore, point }C\text{ is: }\:\left(\frac{184}{\sqrt{197}},\;\frac{212}{\sqrt {197}}\right)$

 April 12th, 2013, 02:56 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,464 Thanks: 2038 By Pythagoras, AB has length $\small5\sqrt{197},$ whereas AC has length 20. The coordinates of C are obtained from those of B by multiplying by AC/AB, which is $\small4/\sqrt{197}.$

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