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1. A) a circle has its centre at the origin and cuts the x-axis at the points A(-c,0) and B(c,0). Write down the equation of the circle.

B) The point P (h,k) lies on the circle. Find expressions for the gradients of AP and BP.

C) By showing that the product of the gradients of AP and BP is -1, prove that <APB= 90

A) x^2 + y^2 =C^2

B) Gradient of AP = K/H+C

C) this is the part im stuck on, I don't know if this is the correct answer I times the gradient of AP and BP to get K^2/H^2-C^2 = -1

By doing that have I proved <APB = 90?
2. (Original post by Vorsah)
A) a circle has its centre at the origin and cuts the x-axis at the points A(-c,0) and B(c,0). Write down the equation of the circle.

B) The point P (h,k) lies on the circle. Find expressions for the gradients of AP and BP.

C) By showing that the product of the gradients of AP and BP is -1, prove that <APB= 90

A) x^2 + y^2 =C^2

B) Gradient of AP = K/H+C

C) this is the part im stuck on, I don't know if this is the correct answer I times the gradient of AP and BP to get K^2/H^2-C^2 = -1

By doing that have I proved <APB = 90?
Well if two lines AP and BP are perpendicular what does it say about angle APB?!
3. (Original post by Mr M)
Well if two lines AP and BP are perpendicular what does it say about angle APB?!
It's 90
4. (Original post by Vorsah)
It's 90
And do you understand that lines are perpendicular if the product of their gradients is -1?
5. (Original post by Mr M)
And do you understand that lines are perpendicular if the product of their gradients is -1?
Yes
6. (Original post by Vorsah)
Yes
So you are happy now?

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Updated: April 7, 2013
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