[aznkid66]

(Original post by krisshP)
Why can't line AB be a chord?

He means perpendicular bisectors.

The perpendicular bisectors of (at least) 2 chords always meet at one point: the center.

OA is a chord whose direction is vertical and whose midpoint is at y=1. Therefore, its perpendicular bisector is the line with the equation y=1.

AB is another chord that's perpendicularly bisected by some line. Do you know how to find the equation for a line that bisects a line segment whose endpoints are given?

[Plato's Trousers]

(Original post by BabyMaths)
It is. You have the equation of the bisector of this chord too.

It's pity you got such bad advice on this.

Answer is (x-8)^2+ (y-1)^2=65

bad advice? How so? It's a perfectly standard method to take three points of a circle and find the centre from them

[krisshP]

(Original post by BabyMaths)
It is. You have the equation of the bisector of this chord too.

It's pity you got such bad advice on this.

Answer is (x-8)^2+ (y-1)^2=65

Bisector equation:


How is the answer (x-8)^2+ (y-1)^2=65?????? Please explain every stage of your working out.

Thanks.

[aznkid66]

(Original post by krisshP)
Bisector equation:


How is the answer (x-8)^2+ (y-1)^2=65?????? Please explain every stage of your working out.

Thanks.

(x-h)^2+(y-k)^2=r^2 is the equation of the circle, where (h, k) is the center.
He's saying the coordinates of the center point is (8, 1).

[blueray]

(Original post by krisshP)
Bisector equation:


How is the answer (x-8)^2+ (y-1)^2=65?????? Please explain every stage of your working out.

Thanks.

When they are saying perpendicalr bisector they mean this.

[BabyMaths]

(Original post by Plato's Trousers)
bad advice? How so? It's a perfectly standard method to take three points of a circle and find the centre from them

You made the question more difficult than it needs to be.

We have y=1 at centre.

Bisector of AB has equation 3y+2x-19=0, sub in y=1 to get x= 8.

(x-8)^2 + (y-1) ^2 =65

r^2 = 65 can be obtained by considering the point (0,0).

[krisshP]

I thought the equation of a circle was:

x^2+y^2=r^2

[blueray]

(Original post by aznkid66)
(x-h)^2+(y-k)^2=r^2 is the equation of the circle, where (h, k) is the center.
He's saying the coordinates of the center point is (8, 1).

(Original post by BabyMaths)
You made the question more difficult than it needs to be.

We have y=1 at centre.

Bisector of AB has equation 3y+2x-19=0, sub in y=1 to get x= 8.

(x-8)^2 + (y-1) ^2 =65

r^2 = 65 can be obtained by considering the point (0,0).

Where did you get y=1 at the center from? Explain this and he will understand.

[Plato's Trousers]

(Original post by krisshP)
I thought the equation of a circle was:

x^2+y^2=r^2

only if its centre is the origin

[krisshP]

(Original post by BabyMaths)
We have y=1 at centre.

How do you know y=1 is the centre so instantly?

[aznkid66]

(Original post by Plato's Trousers)
bad advice? How so? It's a perfectly standard method to take three points of a circle and find the centre from them

Then again, part (a) and part (b) made it pretty obvious what would be in the mark scheme.

Sorry, krisshP, I should've read the question more carefully instead of skimming through everything before (c)

x^2+y^2=r^2 is always centered at (0,0). Think about it in terms of the distance formula r=...

If you know about vertical/horizontal transformations, it's like translating a circle centered at 0 left/right by h and up/down by k.

(Original post by krisshP)
How do you know y=1 is the centre so instantly?

OA is a chord whose direction is vertical (it's on the y axis) and whose midpoint is at y=1 (a vertical line that starts at y=0 and ends at y=2 has a midpoint at y=1).

[BabyMaths]

(Original post by krisshP)
How do you know y=1 is the centre so instantly?

I already explained that. (0,0) to (0,2) is a chord of the circle. Its perpendicular bisector is y=1.

[krisshP]

(Original post by BabyMaths)
I already explained that. (0,0) to (0,2) is a chord of the circle. Its perpendicular bisector is y=1.

ok. I used y=1 and 3y+2x-19 and ended up with y=1 and x=8<--correct answer

Thanks a lot.

[blueray]

(Original post by BabyMaths)
I already explained that. (0,0) to (0,2) is a chord of the circle. Its perpendicular bisector is y=1.

I get the chord part of a circle, but why does that make y=1 if it's a perpendicular bisector? A diagram would help

[aznkid66]

(Original post by blueray)
I get the chord part of a circle, but why does that make y=1 if it's a perpendicular bisector? A diagram would help

Diagram of Point O, Point A (on y-axis), and perpendicular bisector of segment OA.

[blueray]

(Original post by aznkid66)
Diagram of Point O, Point A (on y-axis), and perpendicular bisector of segment OA.

So perpen bisec is half of y value of point A?

[aznkid66]

(Original post by blueray)
So perpen bisec is half of y value of point A?

Yup, it's the equation of all x such that y=1.

[blueray]

(Original post by aznkid66)
Yup, it's the equation of all x such that y=1.

So if A was 0, 10 then y= 5????

[aznkid66]

(Original post by blueray)
So if A was 0, 10 then y= 5????

Yes, you are correct in your reasoning.

[Plato's Trousers]

(Original post by krisshP)
ok. I used y=1 and 3y+2x-19 and ended up with y=1 and x=8<--correct answer

Thanks a lot.

In fact, I am now prepared to admit that I gave you a rather over-complicated way of solving the problem. (I hadn't really read your question properly).

Sorry about that.

Fortunately, better minds than mine were on hand to provide better guidance.

Mea maxima culpa