Grade 10 Math Question?

Question

(The *s means "squared"). The equation of the circle with center 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 center of the circle lies on the right bisector of chord AB

b)Find the distance from the center of the circle to chord AB, to the nearest tenth.

Answer 1

The "right" bisector is perpendicular to AB.

AB has slope ((-2 - 6)/(-6 - -2)) = (-8/-4) = 2

The perpendicular bisector must have slope of the negative reciprocal (-1/2) of the slope of AB.

The perpendicular bisector must also pass through the midpoint of AB, which is the average of the X and Y-values of A and B
((-2 + -6)/2, (6 + -2)/2) =
(-8/2, 4/2) =
(-4, 2)

So... the bisector is a line that has slope -1/2 and passes through (-4,2):

y = (-1/2)x + b
2 = (-1/2)(-4) + b
2 = 2 + b
b = 0

The equation for the bisector is:

y = (-1/2)x

(a) If the center of the circle lies on the right bisector, then the center of the circle will satisfy the bisector's equation:

y =? (-1/2)x
0 =? (-1/2) * 0
0 = 0

b) The shortest distance to AB is the perpendicular distance, and we already have that line (the bisector). You need to find the distance from the center of the circle (0,0) to the midpoint of AB (-4,2).

d = sqrt( (x2 - x1)^2 + (y2 - y1)^2)
d = sqrt( (-4 - 0)^2 + (2 - 0)^2)
d = sqrt(16 + 4)
d = sqrt(20)
d = about 4.5

Answer 2

Eqn of Circle: x² + y² = 40
End points of Chord are A(-2,6) and B(-6,-2)
Ew\qn of Chord is:
[y-6]/[-2-6] = [x-(-2)]/[-6 - (-2)]
(y-6)/(-8)=(x+2)/(-4)
y-6 = 2x+4
y = 2x+10
which is of the form y=m'x +c
Thus slope of the chord is m'=2
For a perpendicular to this line its slope m" has to be
m" = -1/m' = (-1)/(2)=-1/2

Mid point of the chord is given by
x'=(-6-2)/2=-4
y'=(6-2)/2=2
Mid point F is (-4,2)

A line passing through F(-4,2) with slope= -1/2 is given by
y=m"x +c, where c is the y-intercept
2=(-1/2)(-4) + c
c=0

Therefore, equation of the right bisector of the chord is
y=(-1/2)x
2y + x = 0

The coordinate of the center satisfies the equation, so the center lies on the right bisector of chord AB........(a)

Since DE is the right bisector of the chord the distance from the center of the circle to the chord AB is equal to OF
Distance OF = √[(-4-0)² +(2-0)²]
=√(16+4)
=√20
=4.4721
=4.5 units

Answer 3

a) There are different ways you could prove this. One I suppose is to find the equation for line DE and show that it passes through (0,0). The slope of line AB is (-2-6)/(-6-(-2)) = -8/-4 = 2. So the slope of DE is -1/2 because it's perpendicular. It includes the midpoint of AB, which is ((-2-6)/2, (6-2)/2) = (-4,2). Use this and the slope to find the equation for the line, and verify that (0,0) is on the line.

b) This is the distance from the center to the midpoint of AB. You already know both, so the center should be easy.

Answer 4

Bisector Of A Circle

Answer 5

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