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Circle geometry help

The points A(1,3), B(1,7), C(7,7) lie on the circle (x4)2+(y5)2=13 (x-4)^2 + (y-5)^2 = 13
a) draw a sketch
b) by considering the gradients on AB and BC show that AC is a diameter of the circle


for part (a) I've done something that resembles this: http://gyazo.com/7f2b0d1853cc5d8580ba2ea89eff17d2

and for part (b)
Gradient of AB = 40\frac{4}{0} (infinity?)
Gradient of BC = 0
not sure how this will help me to show that AC is the diameter? I can see of the sketch that AC is the diameter but how do I use the gradients to show this?

Thanks!
Reply 1
Original post by Secret.
The points A(1,3), B(1,7), C(7,7) lie on the circle (x4)2+(y5)2=13 (x-4)^2 + (y-5)^2 = 13
a) draw a sketch
b) by considering the gradients on AB and BC show that AC is a diameter of the circle


for part (a) I've done something that resembles this: http://gyazo.com/7f2b0d1853cc5d8580ba2ea89eff17d2

and for part (b)
Gradient of AB = 40\frac{4}{0} (infinity?)
Gradient of BC = 0
not sure how this will help me to show that AC is the diameter? I can see of the sketch that AC is the diameter but how do I use the gradients to show this?

Thanks!


Not sure if that question is right. But try considering the gradients OA and OC instead, where O is the origin of the circle
Reply 2
Original post by 2710
Not sure if that question is right. But try considering the gradients OA and OC instead, where O is the origin of the circle



OA = 3 and
OC = 1?
Reply 3
Original post by Secret.
OA = 3 and
OC = 1?


O is the centre of the circle, not (0,0). What is the centre of the circle?
Reply 4
Original post by 2710
O is the centre of the circle, not (0,0). What is the centre of the circle?



Do I work that out from the midpoint of AC?
Reply 5
Original post by Secret.
Do I work that out from the midpoint of AC?


You should have learnt this?

(x4)2+(y5)2=13 (x-4)^2 + (y-5)^2 = 13

The centre of the circle is just (4,5). -1 times the number in the x bracket, and -1 times the number in the y bracket.
Reply 6
Original post by 2710
You should have learnt this?

(x4)2+(y5)2=13 (x-4)^2 + (y-5)^2 = 13

The centre of the circle is just (4,5). -1 times the number in the x bracket, and -1 times the number in the y bracket.


Omg Yes! Sorry, I completely forgot the equation was even there!
But how come the question suggests working out the gradients of AB and AC?
Reply 7
Original post by Secret.
Omg Yes! Sorry, I completely forgot the equation was even there!
But how come the question suggests working out the gradients of AB and AC?


Is this on a paper? Or did your teacher give it to you. Maybe a typo or something? It cannot be AB because as u said, the gradient of that is infinity
Reply 8
Original post by 2710
Is this on a paper? Or did your teacher give it to you. Maybe a typo or something? It cannot be AB because as u said, the gradient of that is infinity


From a textbook
and I think what that means is it is a verticle straight line, which it is according to the diagram I made (the line x=1)
(edited 11 years ago)
Reply 9
A straight vertical line is perpendicular to a straight horizontal line =_="

If you have a right triangle circumscribed by a circle, what is the triangle's hypotenuse with respect to the circle?
Reply 10
Original post by aznkid66
A straight vertical line is perpendicular to a straight horizontal line =_="

If you have a right triangle circumscribed by a circle, what is the triangle's hypotenuse with respect to the circle?


Oh yeh lol
Reply 11
Original post by aznkid66
A straight vertical line is perpendicular to a straight horizontal line =_="

If you have a right triangle circumscribed by a circle, what is the triangle's hypotenuse with respect to the circle?


The diameter? I said that above but I thought it wanted working out using the gradients of AB and AC
Reply 12
Original post by Secret.
The diameter? I said that above but I thought it wanted working out using the gradients of AB and AC


One of the circle theorems states that the angle formed by a triangle in a circle when the hypotenuse is the diameter is a right angle. show that the two lines are perpendicular and that the hypotenuse passes through the centre of the circle.

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