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Circle Geometry

Q: 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.

The point P(h, k) lies on the circle. Find expressions for the gradients of AP and BP. By showing that the product of the gradients of AP and BP is -1, prove that angle APB = 90.

Help would be really appreciated.

This is what I've done so far:
image.jpg
As you can see, it is the second part of the question that I am stuck on.
Ok, first let's get the gradients of both AP and BP, which respectively are k/(h+c) and k/(h-c), now, think how h and k are related(the circle formula).
Original post by Argylesocksrox
Ok, first let's get the gradients of both AP and BP, which respectively are k/(h+c) and k/(h-c), now, think how h and k are related(the circle formula).


Are you referring to (x-x1) + (y-y1) = r^2?
Original post by TarotOfMagic
Are you referring to (x-x1) + (y-y1) = r^2?


No, the fact that x^2+y^2=c^2, given that H and K both lie on the circle, they satisfy the equation h^2+k^2=c^2

given that I already gave you the gradients of AP and BP, the algebra should fallout when you multiply AP and BP
Original post by Argylesocksrox
No, the fact that x^2+y^2=c^2, given that H and K both lie on the circle, they satisfy the equation h^2+k^2=c^2

given that I already gave you the gradients of AP and BP, the algebra should fallout when you multiply AP and BP


The two gradients multiplied together gives k^2/h^2-c^2, not -1 though?
Original post by TarotOfMagic
The two gradients multiplied together gives k^2/h^2-c^2, not -1 though?


How may h^2-c^2 be rewritten? (circle formula)
Original post by Argylesocksrox
How may h^2-c^2 be rewritten? (circle formula)


h^2 + k^2 = c^2?
Original post by TarotOfMagic
h^2 + k^2 = c^2?


it may be written as -k^2.
Original post by Argylesocksrox
it may be written as -k^2.


But how would I prove that this is perpendicular and 90 degrees?
Close thread. Question solved, thanks for helping!

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