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The question in question (ha!)

I'm struggling with the above question. After completing:
(i) radius = $\sqrt{13}$, centre = $(-4,2)$;
(ii) A=$(-7,0)$, B=$(-1,0)$;
(iii) has left me stumped.

I can't see any way to find the points without going through the entire proof of "chords perpendicular to a radius are bisected by the radius", and that seems excessive considering the calculations I have already done below take up a substantial amount of the 7 marks.

I know that line DE has the equation $y = 2x + 6$ and the lines cross at $(-2.2,1.1)$, but I'm not quite sure how to progress from here.

How would I find points D and E?

Corsaka
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1 Answers1

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You know the equation of the circle. You also know the equation of the line DE. Solving the simultaneous gives a quadratic, which will give you the two solutions for the two intercepts D and E.

I suspect this is the method that may be expected, although perhaps there is also a cleverer way. This solution uses (i) of the question, and if my memory serves, usually these problems try to guide you through them by making you use facts you have already established. So I don't think this method would be excessive for the given marks.

masiewpao
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    Immediately before (i) the question gives the formula for the circle, so there is no need to calculate the formula. Just plug in $y=2x+6$ and solve the resulting quadratic. You are right on the money with the approach. I doubt there is a more straightforward approach. – SlipEternal Mar 03 '20 at 11:37
  • @InterstellarProbe oh my mistake, I will fix that. Yes I agree, it seems like quite a standard approach. – masiewpao Mar 03 '20 at 11:56
  • Ah, so the answer is: I'm an idiot. Thank you! – Corsaka Mar 03 '20 at 12:00
  • @Corsaka Not at all, sometimes you just need another perspective! – masiewpao Mar 03 '20 at 12:10