If a diameter has endpoints (a,b) and (c,d), prove that the equation of the circle can be written $(x-a)(x-c)+(y-b)(y-d)=0$

Question

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.

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?

Answer 1

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]

Answer 2

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.)

Answer 3

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.