Standard form of the equation of a circle : does the order matter?

Question

The distance formula is

$d = \sqrt{(x2 - x1)² + (y2 - y1)²}$

If the two points are (1, 2) (3, 4), it doesn't matter whether one write :

$d = \sqrt{(1 - 3)² + (2 - 4)²}$

Or

$d = \sqrt{(3 - 1)² + (4 - 2)²}$

Now the standard form of the equation of a circle is said to be :

$(x - h)² + (y - k)² = r²$

Could it also be :

$(h - x)² + (k - y)² = r²$

I'd instinctively say "yes". But if I test with $r = \sqrt{20}$, the center $(-1, 2)$, and $x = 3$, I somehow can't get to the same answer with both versions of the equation.

My mind is probably stuck on something stupid...

EDIT:

Alright so both possibilities are:

$(x + 1)² + (y - 2)² = 20$

And

$(-1 - x)² + (2 - y)² = 20$

For x = 3, first version :

$$(3 + 1)² + (y - 2)² = 20$$ $$(y - 2)² = 20 - 16$$ $$y - 2 = 2$$ $$y = 4$$

and second version :

$$(-1 - 3)² + (2 - y)² = 20$$ $$(2 - y)² = 20 - 16$$ $$2 - y = 2$$ $$-y = 2 - 2$$ $$y = 0$$

Or maybe it's just time to go to sleep?

Answer 1

No the order doesn't matter.

In general $(a-b)=-1\cdot (b-a)$ so $(a-b)^2=(-1)^2(b-a)^2=(b-a)^2$.

So $(x-h)^2=(h-x)^2$ for all $x,h$.

You jump from $(2-y)^2=4$ to $2-y=2$, but $(-2)^2=4$, also, so you really have $2-y=\pm 2$ or $y=0\text{ or } 4$.

And similarly for $(y-2)^2=4$, then $y-2=\pm 2$ or $y=0\text{ or }4$.

Answer 2

Prove that for $f(x) = x^2$ we have $f(x) = f(-x)$ or in other words it's even function. It shouldn't be hard at all.

Answer 3

You simplified by square rooting both sides. Square roots don't cancel squares; plus or minus square roots cancel squares. Do a plus or minus square root on both sides and you'll get matching responses.

Best, Kevin