the equation of a circle with centre C(0,0) is x squared + y squared = 100. the points A(8,-6) and B(-10,0) are the endpoints of chord AB.

Question

DE right bisects the chord AB at F. Verify that the centre of the circle lies on the right bisector of chord AB.

Answer 1

There are three parts to this problem. First we find the slope of AB. Since DE is the perpendicular bisector of AB, its slope will be the negative reciprocal of the slope of AB.

Then we find the midpoint of AB, which is called F in the problem. DE will be the line with that slope that passes through the point F.

Finally, we need to show that DE passes through the Origin since that is the center of the circle.

 

Part 1 - Find the slope of AB.  Use the slope formula   (y₂-y₁)/(x_2-x₁) = (-6 - 0)/(8 - -10) = -6/18 = -1/3.

 

    The slope of DE is the negative reciprocal of -1/3, so its slope is  3

 

Part 2 - Find the midpoint of AB.  Use the midpoint formula, which is just the average of the two x-coordinates and the average of the two y-coordinates

 

    x =  (-10+8)/2 =  -2/2 = -1               y = (0+ -6)/2 = -6/2 = -3    So point F is   (-1, -3)

 

Part 3 - Find the line with slope  3  that passes through the point  (-1, -3)

 

   Use the slope-intercept formula    y - y₁ = m(x - x₁)

 

     y - (-3) = 3(x - (-1))

 

     y + 3 = 3(x + 1)

 

We don't have to clean up the equation, because all we need it for is to verify that it passes through the Origin, which is (0,0). We verify this by replacing both x and y by 0, and seeing that the equation is true.

 

     0 + 3 = 3(0 + 1)

 

           3 = 3

 

Yes, it is true, so we have proven what was asked.