Question 924004
based on the following, your radius has to be perpendicular to the chord.


<a href = "http://www.regentsprep.org/regents/math/geometry/gp14/circlechords.htm" target = "_blank">http://www.regentsprep.org/regents/math/geometry/gp14/circlechords.htm</a>


so you get a triangle formed by the radius of the circle and the chord such that the ends of the chord that connect to the center of the circle are each a radius of the circle.


the chord is 4 units in length.


two right triangles are formed with each right triangle having a base of 2 which is 1/2 the length of the chord.


the length of the line segment formed by the radius that bisects the chord is found by taking the 2 points common to that radius and finding the distance between them.


the two points are (-3,-2) and (3,1)


the distance between them will be square root of [(-2-1)^2 + (-3-3)^2] which becomes square root of [9 + 36] which becomes square root of (45).


your right triangle has one leg of 2 and one leg of sqrt(45)


to find the hypotenuse of the right triangle which is the radius of the circle, you need to use the pythagorean formula.


you will get r^2 = 2^2 + sqrt(45)^2


that becomes r^2 = 4 + 45 which makes r^2 = 49 which makes r = 7.


the radius of your circle has to be 7.


the equation of your circle becomes:


(x+3)^2 + (y + 2)^2 = 49


a graph of the circle and the chord and the radius that bisects the chord is shown below:


<img src = "http://theo.x10hosting.com/2014/111605.jpg" alt="$$$" </>


all the numbers check out.


the radius is 7
each sice of the chord that is bisected by the radius has a length of 2.