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# Circle C1 question watch

1. So was doing this past paper and got full marks up until this question. I'm not really too sure how to do ithttp://www.ocr.org.uk/download/pp_07_jan/ocr_17592_pp_07_jan_l_gce.pdf

10 part 2
2. (Original post by hazbaz)
So was doing this past paper and got full marks up until this question. I'm not really too sure how to do ithttp://www.ocr.org.uk/download/pp_07_jan/ocr_17592_pp_07_jan_l_gce.pdf

10 part 2
You have an equation of a circle and you know an x value and a y value. Substitute in and you form a quadratic in k. Only one of the solutions fits the condition given in the question.
3. (Original post by hazbaz)
So was doing this past paper and got full marks up until this question. I'm not really too sure how to do ithttp://www.ocr.org.uk/download/pp_07_jan/ocr_17592_pp_07_jan_l_gce.pdf

10 part 2
Sub x=-3 into the equation of the circle and solve for y. Since k<0, you want the negative y value.
4. (Original post by hazbaz)
So was doing this past paper and got full marks up until this question. I'm not really too sure how to do ithttp://www.ocr.org.uk/download/pp_07_jan/ocr_17592_pp_07_jan_l_gce.pdf

10 part 2
Let's say (A,B) is the centre of the circle. (-3,k) is a point on the circle. Hence the distance between those two points is the radius.

You know the radius of the circle from part (i).

r2 = (x2-x1) + (y2-y1), where r is the radius. (this is just applying Pythagoras' theorem).

Then rearrange to find k.
5. (Original post by Mr M)
You have an equation of a circle and you know an x value and a y value. Substitute in and you form a quadratic in k. Only one of the solutions fits the condition given in the question.
Well that was easier then i expected thanks. I think when i see the number of marks on some of these questions i panic and think loads of stuff needs to be done.

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