# OCR C1 CirclesWatch

#1
I was trying a past paper and I don't understand how to do this question.

The point A (2,2) lies on the circumference of C. Given that AB is a diameter of a circle, find the coordinates of B.

The radius is square root 40 and the answer given in B=(-2,-10).
What do you have to do to get the answer?
0
4 years ago
#2
I was trying a past paper and I don't understand how to do this question.

The point A (2,2) lies on the circumference of C. Given that AB is a diameter of a circle, find the coordinates of B.

The radius is square root 40 and the answer given in B=(-2,-10).
What do you have to do to get the answer?
You have to give us all of the question
0
#3
(Original post by TenOfThem)
You have to give us all of the question
http://i.imgur.com/dj32vAs.png
0
4 years ago
#4
You know where A is and you know where C is

And you know that AC = CB

In other words C is the mid-point of AB
0
#5
(Original post by TenOfThem)
You know where A is and you know where C is

And you know that AC = CB

In other words C is the mid-point of AB
Isn't C the circumference?
0
4 years ago
#6
Have you found the centre of the circle from part i.?

The centre is the midpoint of AB
0
4 years ago
#7
Isn't C the circumference?
Sorry I meant to indicate that C was the centre
0
#8
(Original post by Jordan97)
Have you found the centre of the circle from part i.?

The centre is the midpoint of AB
yeah, (0,-8)
0
4 years ago
#9
yeah, (0,-8)
Your y co-ordinate is wrong. What did you do to find it?
0
#10
(Original post by Jordan97)
Your y co-ordinate is wrong. What did you do to find it?
completed the square so x^2+(y-(-8))^2 which is x^2+(y+8)^2
0
4 years ago
#11
completed the square so x^2+(y-(-8))^2 which is x^2+(y+8)^2
Half of 8 is 4
0
#12
(Original post by TenOfThem)
Half of 8 is 4
oh yeah, thanks
0
4 years ago
#13
completed the square so x^2+(y-(-8))^2 which is x^2+(y+8)^2
When completing the square you half the coefficient of the linear variable so y^2+8y = (y+4)^2
0
#14
(Original post by Jordan97)
When completing the square you half the coefficient of the linear variable so y^2+8y = (y+4)^2
yeah, thanks for reminding me. Do I find the gradient next?
0
4 years ago
#15
yeah, thanks for reminding me. Do I find the gradient next?
It's really simple from here. The centre is the midpoint of AB, so find the distance between the x and y co-ordinates of A and the centre, double it and add that to A co-ordinates.
1
#16
(Original post by Jordan97)
It's really simple from here. The centre is the midpoint of AB, so find the distance between the x and y co-ordinates of A and the centre, double it and add that to A co-ordinates.
oh wow, did not think of that. Thanks for helping me
0
4 years ago
#17
oh wow, did not think of that. Thanks for helping me
haha no worries!
0
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