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# GCSE MATHS HELP circles and triangles watch

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1. A circle has the equation x^2+y^2=80.
The enter of the circle is the origin, O.
The point P on the circle has the coordinates (2a,a) ere a is a positive constant.
The tangent to the circle at P crosses the x axis at point Q and the y axis at point R.
Work out the area of triangle OQR
2. (Original post by EVIEDOBSON_)
A circle has the equation x^2+y^2=80.
The enter of the circle is the origin, O.
The point P on the circle has the coordinates (a,2a) where a is a positive constant.
The tangent to the circle at P crosses the x axis at point Q and the y axis at point R.
Work out the area of triangle OQR
Substitute (a,2a) into x^2+y^2=80 so a^2 +4a^2=80 a^2=80/5=16 a=4 since positive constant.
The triangle is a right angle triangle (Try to sketch it). so use area of a triangle= half base times height. base is the x-coordinate of p and height is y-coordinate of p. so area=1/2*4*8=16. Hope this helps!
3. (Original post by anujsr)
Substitute (a,2a) into x^2+y^2=80 so a^2 +4a^2=80 a^2=80/5=16 a=4 since positive constant.
The triangle is a right angle triangle (Try to sketch it). so use area of a triangle= half base times height. base is the x-coordinate of p and height is y-coordinate of p. so area=1/2*4*8=16. Hope this helps!
Isn't that just the triangle up to p? Because we're looking for OQR Attachment 617106
4. (Original post by EVIEDOBSON_)
Isn't that just the triangle up to p? Because we're looking for OQR Attachment 617106
Sorry about that. I didn't read the whole of the question. The approach is the same up to getting the value of a. The gradient of the line OP is just 2a/a. which is 2. From gradient of a perpendicular line, the gradient of the line RQ is -1/2.(m1*m2=-1). find the equation of the line RQ using the fact that m=-1/2 and P lies on that line. Using this line find the x-intercept(Q) and y-intercept(R) now apply half base times height which is 1/2 * x-intercept(R) * y-intercept. Hope this helps!!

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Updated: February 4, 2017
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