Co-ordinate Geometry

L

lovely_nancy

New member
Joined
Sep 14, 2010
Messages
29
I Have a question I solved some of it but kinda stuck
and now need someone's help :S

A(6,6), B(6,-2), C(-1,-1) and D(-2,2) are four points on a circle.
first it asks me to find the equations of perpendicular bisector of AB and CD which I have done
but now it asks me to:
Hence find the co-ordinates of the center of the circle

do i find the midpoint of AB to give me the centre or something else?
 
A(6,6), B(6,-2), C(-1,-1) and D(-2,2) are four points on a circle.
first it asks me to find the equations of perpendicular bisector of AB and CD which I have done
but now it asks me to:
Hence find the co-ordinates of the center of the circle
Click to expand...

Think about the circle that these four points lie on. AB and CD are chords on that circle. The perpendicular bisectors of chords always pass through the center of the circle. Right? Therefore, the intersection of two perpendicular bisectors will be the center of the circle.
 
if im nt wrong then do I use the equations I found
and use simultanious equation to get the two points??
 
if im nt wrong then do I use the equations I found
and use simultanious equation to get the two points??
Click to expand...

Yes, you will need to find the equations for the lines that are the perpendicular bisectors. You will find one point from this system of equations. That point is defined by one x value and one y value.
 
The equations I found are
y=-3x -4
y= -8x + 54

how do I put them into simultanious equation form?? :S
 
yes this is the graph i came up with but thought I was wrong
im really confused on how to find the co-ordinates now
 
Perahps a little complicated, but you could use the circle equation.

\(\displaystyle (-2-a)^{2}+(2-b)^{2}=r^{2}\)

\(\displaystyle (6-a)^{2}+(6-b)^{2}=r^{2}\)

\(\displaystyle (-1-a)^{2}+(-1-b)^{2}=r^{2}\)

Three variables with three unknowns. a and b give the center cooridnates and r is the radius.

Set the first two equal and solve for a.

\(\displaystyle a=\frac{8-b}{2}\)

Sub into the third and first and set equal:

\(\displaystyle \frac{5}{4}b^{2}-3b+26=\frac{5}{4}b^{2}-10b+40\)

Solve for b and a follows. As does r.
 
I pretty much just gave it to you. After you find a and b (the center), just use \(\displaystyle (x-a)^{2}+(y-b)^{2}=r^{2}\)
 
lovely_nancy said:
The equations I found are

y = -3x - 4 This is not the perpendicular bisector of chord CD.

It's the equation of the line passing through points C and D.



y = -8x + 54 I don't know from where this equation comes.

how do I put them into simultanious equation form? See below.
Click to expand...

NOTE: The chords plotted on galactus' graph are not correct.

By inspection, chord AB is clearly vertical with midpoint (6, 2). This tells us that the perpendicular bisector has slope zero.

Therefore, the equation of chord AB's perpendicular bisector must be:

y = 2

Your equation above shows that chord CD has a slope of -3.

Therefore, the perpendicular slope is 1/3.

The midpoint formula gives (-3/2, 1/2).

Using these three values in the Point-Slope formula gives the equation of CD's perpendicular bisector:

y = x/3 + 1

Since y is the same on each of the perpendicular bisectors at their intersection point, we solve the two equations simultaneously by setting the expressions for y equal to one another.

2 = x/3 + 1

Clearly, the x-coordinate at the center of the circle is 3.

To find the corresponding y-coordinate, we substitute x = 3 into y = x/3 + 1 and evaluate to get 2.

The coordinates at the center of the circle are (3, 2).

 
Oops, I sure did. :oops:

Sorry about that. I clearly connected up the wrong points.

I went about the calculations another way, but we got the same there.

I arrived at the equtions \(\displaystyle -3b+10=14, \;\ \frac{8-b}{2}=a\)

a=3 and b=2.

I had a nice looking circle and then went and screwed it up by connecting up the wrong points.

Thanks for catching it though. Frankly, I didn't even notice that part of the problem.

I just saw the points and find the center. That's what I get for not paying attention.
 
galactus said:
I went about the calculations another way

And a lovely way it is, too.
Click to expand...

I followed-up on wjm11's information to Nancy because it appears that Nancy veered off-course, on that approach.

 
Yes, mm, had I paid attention I would have noticed the easier way.

The method I used works, but why dig a hole with a spoon when you can use a backhoe.

To redeem myself, though it may be a moot point, here is the circle.

The bisectors can be seen to intersect at the center.

\(\displaystyle \frac{x}{3}+1=2\)

Can I get out of the corner now?. SK, you can come with me :D
 
SK does not want to leave the corner: his wife is waiting to send him to the doghouse :shock:
 
You must log in or register to reply here.
Top