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 ?
The equation x^2 + y^2 = 40 represents the circle with center O(0,0) and radius √40. In this problem, we are given that the points A(-2,6) and B(-6,-2) are endpoints of chord AB. To understand how the equation of the circle relates to the problem, let's break it down step by step:
a) To verify that the center of the circle lies on the right bisector of chord AB, we need to find the equation of the right bisector and check if the coordinates of the center satisfy this equation.
The right bisector is a line that passes through the midpoint of chord AB (let's call it M) and is perpendicular to AB. To find M, we first need to find the coordinates of the midpoint.
The midpoint formula states that the coordinates of the midpoint are the average of the coordinates of the endpoints. Thus, for AB, the midpoint M is:
M = ((-2+-6)/2, (6+-2)/2)
= (-4, 2)
Now that we have the midpoint M, let's find the equation of the right bisector. Since AB's slope is (change in y / change in x), we can find the perpendicular slope by taking the negative reciprocal of the slope of AB. The negative reciprocal of AB's slope is (1/AB's slope).
AB's slope = (change in y / change in x) = (-2-6)/(-6-(-2)) = -8/-4 = 2
Perpendicular slope = 1/AB's slope = 1/2
Using the point-slope form of a line, where y - y1 = m(x - x1), we can plug in the values of M and the perpendicular slope to get the equation of the right bisector:
y - 2 = (1/2)(x + 4)
Simplifying this equation, we get:
2y - 4 = x + 4
x - 2y = -8
To verify that the center of the circle lies on the right bisector, we substitute the coordinates of the center O(0,0) into the equation of the right bisector:
0 - 2(0) = -8
0 = -8
Since the equation holds true, we can conclude that the center of the circle, O(0,0), lies on the right bisector of chord AB.
b) To find the distance from the center of the circle to chord AB, we can use the formula for the distance between a point and a line.
The distance between a point (x1, y1) and a line Ax + By + C = 0 is given by:
d = |Ax1 + By1 + C| / √(A^2 + B^2)
In this case, the equation of the right bisector is x - 2y = -8, which can be rewritten as x - 2y + 8 = 0.
Comparing this equation with Ax + By + C = 0, we have A = 1, B = -2, and C = 8.
Substituting the coordinates of the center O(0,0) into the formula, we get:
d = |(1)(0) + (-2)(0) + 8| / √((1)^2 + (-2)^2)
= |0 + 0 + 8| / √(1 + 4)
= 8 / √5
≈ 3.6 (rounded to the nearest tenth)
Therefore, the distance from the center of the circle to chord AB is approximately 3.6.