C2 - Circle Geometry

So im doing this mixed exercise with 20 circle geometry questions, and when i try this question algebraically i get an infinite gradient ill let you have alook at this.

The points A (-7,7), B(1,9), C(3,1) and D(-7,1) lie on a circle. The Lines AB and CD are chords of the circle.

a) Find the equation of the perpendicular bisector of i)AB ii)CD
b) Find the coordinates of the centre of the circle

a) A(-7,7) B (1,9) - Find gradient of AB

m = 9-7/ 1- (-7) = 2/8 = 1/4 therefore the gradient for the perpendicular bisector = -4

A perpendicular bisector gets the midpoint of the chord AB.

(-7+1/2, 9+7/2) = (-3,8) -- Now we have a point where the perpendicular bisector insects and a gradient enough info to find the equation of the perpendicular bisector

y-y1 = m(x - x1) - Equation for finding the equation of a line

y - 8 = -4(x + 3)
y= -4x - 4 --- Equation of Perpendicular bisector of AB

Ive explained what i did for the first equation, so ill just do the maths now for CD.

C(3,1), D(-7,1)
m = 1-1/-7-3 = 0/-10 ------ uh ohhh Perpendicular = 10/0 = infinity

so m = infinity

Mid point = (-2,1)

Again we have a point and a gradient

y - y1 = m(x - x1)

y - 1 = infinity(x+2)

y -1 = infinity(x) + 2(infinity)

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

y - 1 = x/0 + 2/0 - multiply both sides by zero

(y)0 - (1)0 = x + 2
x = -2

Its the right answer and if you look at the points graphically you can see the perpendicular bisector is x = -2. I was Just wondering can i manipulate infinity like this. Bearing in mind i have never done anything on infinity except the fact that i know 1/0 = infinity.

Reply 1

ebinsc

You have the correct answer, but don't mess around with infinity.

The following method is much easier:

(-7,1) and (3,1) are on the same line y=1

Midpoint of the points = (-2,1)

Therefore perpendicular bisector is x=-2

Hope this helps :smile: