posted by Cynthia on .
The equation of a circle with centre O(0,0) is x^2+y^2=40. The points A(-2,6) and B(-6,-2) are endpoints of chord AB. DE right bisects chord AB at F.
a)Verify that the centre of the circle lies on the right bisector of chord AB.
b)Find the distance from the centre of the circle to chord AB, to the nearest tenth.
now, i can do the problem, but i just don't understand where i use x^2+y^2=40.. i mean what's the purpose of it??? PLEASE HELP ME!!
From the given circle equation you would have to know that its centre is (0,0) and its radius is √40
Notice that both A and B satisfy the equation of the circle, so A and B lie on the circle and AB is indeed a chord.
In a) you are probably expected to illustrate the property that the right-bisector of any chord passes through the centre of the circle.
You claim you did that, ok.
BTW, what did you get for the distance from the centre to chord AB ?