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I am faced with the question in the title and have shown it on my own, but my friend has been asking how to solve it and I want to see if there is a quicker method than what I have used.

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I stopped because I originally expanded out all of the brackets and collected terms together but it gets very method, so I was wondering if you could fairly quickly factorise the expression into the desired form in the title?

Jamminermit
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  • I am surprized you haven't selected @Stinking Bishop solution as the best solution. – Jean Marie Sep 10 '19 at 14:19
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    Yeah, unfortunately although his solution may be technically the best one, it is not using techniques that I or my friend fully understand, so the answer I selected best represents what I was looking for. I do acknowledge the other solution as being neat, and if you believe it would benefit others with a similar question in the future I could select it. – Jamminermit Sep 10 '19 at 17:49
  • I appreciate your very nuanced answer. – Jean Marie Sep 10 '19 at 18:19

3 Answers3

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The equation is simply the statement that the vectors $(x-a, y-b)$ and $(x-c, y-d)$, i.e. $\vec{ZP}$ and $\vec{ZQ}$ are orthogonal. (Use dot product.)

  • From what others posters are saying, this seems to be the neatest method, however I am unfamiliar with the dot product, could you possibly detail the method more? – Jamminermit Sep 09 '19 at 19:11
  • @Jamminermit There is a lot to it, more than I can fit into this comment. You may want to start with https://en.wikipedia.org/wiki/Dot_product . –  Sep 10 '19 at 08:41
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You have an easier job using Pythagoras on the sides of the triangle:

$$\left[(x-a)^2+(y-b)^2\right]+\left[(x-c)^2+(y-d)^2\right]=(a-c)^2+(b-d)^2$$

or (noting that the squares of constants cancel and dividing though by $2$)

$$x^2-ax-cx+y^2-bx-dx=-ac-bd$$

from which the conclusion follows.

[But the other way using the scalar/dot product is neater and shows more insight]

Mark Bennet
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  • How do we know the squares of the constant terms cancel? – Jamminermit Sep 09 '19 at 19:01
  • @Jamminermit What I mean is that the $a^2$ on the left when you expand is clearly going to cancel with the $a^2$ on the right and likewise for $b^2, c^2, d^2$: if you see it right, it is obvious that they appear once on each side. On the other hand you get $2x^2$ and $2y^2$ because the terms with those squares are on the same side - and the cross terms in the expansions all have a factor $2$ as well. – Mark Bennet Sep 09 '19 at 21:05
  • Yeah I see now thanks. I thought you meant was that the -ac and -bd cancelled which got me a but confused, but I understand now. – Jamminermit Sep 10 '19 at 05:32
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Upon completing the squares we get $$(x-a)(x-c)+(y-b)(y-d)=0 \iff$$

$$ x^2-(a+c)x+ac +y^2-(b+d)y+bd=0\iff$$

$$(x-\frac {(a+c)}{2})^2+(y-\frac {(b+d)}{2})^2 =\frac {(a-c)^2}{4}+\frac {(b-d)^2}{4}=R^2$$

Where $$R=\frac {\sqrt {(a-c)^2+(b-d)^2}}{2}$$

That is the equation of the described circle.

J. W. Tanner
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Mohammad Riazi-Kermani
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