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Is there a quicker way to do this question? (proving circle equation)

Question (part b): http://prntscr.com/rsedx3
Midpoint is (a/2 + c/2, b/2 + d/2) and the radius is the distance from P to M so the circle equation would be:

(x - (a/2 + c/2))^2 + (y - (b/2 + d/2))^2 = 1/4(a^2 + b^2 + c^2 + d^2) - 1/2(ac + bd)

I would then expand the brackets and move everything to the LHS & try to complete the square but in an exam this would probably take too long to do & I might make a mistake. Is there an alternative approach?

Reply 1

RDKGames

Study Forum Helper

Original post by TSR360

Yes... it does say 'hence' which you should try to think about.

Part (a) simply wants you to realise that there is a right-angled triangle with the hypotenuse being the diameter PQ.

The point Z is where the right-angle occurs. The fact that Z is any point on the circle gives you an important property you should seek to explore;

ZP^2 + ZQ^2 = PQ^2

by simple Pythagoras.

ZP^2 = (x-a)^2 + (y-b)^2

ZQ^2 = (x-c)^2 + (y-d)^2

PQ^2 = (a-c)^2 + (b-d)^2

Thus, the eqn becomes

(x-a)^2 + (y-b)^2 + (x-c)^2 + (y-d)^2 = (a-c)^2 + (b-d)^2

Move the RHS to LHS and factorise things together. Correct factorisation can be noticed by looking at the answer and noting that x,a,c should be all together as a grouping, and y,b,d should be all together as another grouping.