# Homework Help: : Centre Point of a circle

1. Oct 31, 2007

URGENT: Centre Point of a circle

1. The problem statement, all variables and given/known data
Find the center point of the circle shown in the diagram below and the radius of the circle.

http://img142.imageshack.us/img142/9469/45409301jg6.png [Broken]

The red dots have the following coordinates; (-17,0); (0,17); (31,0)

2. Relevant equations
$$\left( {x + a} \right)^2 + \left( {y + b} \right)^2 = r^2$$

3. The attempt at a solution
I can find the x-coordinate of the center of the circle finding the center point between (-17,0) and (31,0), and as the circle is symmetrical, this will be the x-coordinate of the center point. The value is 7, therefore the center of the circle is located (7,c) where c is some constant.

how do i find the value of the constant?

Many thanks,

Last edited by a moderator: May 3, 2017
2. Oct 31, 2007

### learningphysics

The distance between (7,c) and (0,17) equals the distance between (7,c) and (31,0).

use this to find c, and and then r also...

3. Oct 31, 2007

sorry i don't see how that helps

would simultaneous solving of some sort help?

Last edited: Oct 31, 2007
4. Oct 31, 2007

### HallsofIvy

The perpendicular bisector of any chord in a circle goes through the center. You used that when you calculated the midpoint of the horizontal line between (-17,0) and (31,0): its midpoint is (7,0) and its perpendicular bisector is x= 7. Now choose another two of those points, say (-17, 0) and (0, 17). The midpoint of that segment is (-17/2, 17/2). The slope of that segment is (17-0)/(0-(-17))= 1. The slope of a line perpendicular to that is -1. The equation of the perpendicular bisector of that segment is y= -x+ 17. The center is at the intersection of y= -(x+ 17/2)+ 17/2 and x= 7. That should be easy to solve.

5. Oct 31, 2007

### Smartass

So the centre is (7, c).
Length of segment joining (-17, 0) and center
r^2 = 24^2 + c^2
r^2 = 576 + c^2 _____(Eq 1)
Length of segment joining (0, 17) and center
r^2 = 7^2 + (c-17)^2
r^2 = 49 + c^2 + 289 -34c _______(Eq 2)

Comparing Eqs. 1 and 2:
576 + c^2 = 49 + c^2 + 289 -34c
c = -7

Therefore the centre of the circle is (7, -7).

Finding radius now is pretty trivial.

6. Oct 31, 2007

cheers for that, makes sense now, the radius is $$\sqrt {149}$$? correct...