What is the significance of the circle equation $x^2 + y^2 = r^2$?

140 Views Asked by At

I can’t post images but imagine a circle centered on $0$ with radius $5$ and a point at $(3,3)$ and a point at $(4,4)$.

I know that this is a circle equation but there something I can’t figure out. If I want the distance between the point $(4,4)$ and the circle, I would use Pythagorean’s theorem and so the distance is $\sqrt{4^2+4^2} - r$ right? But there is another equation that gives a number equal to $0$ on the circle, increases positively as we go far away from the circle and increases negatively as we step inside the circle. This equation is $x^2+y^2=r^2$. With the point $(4,4)$, we get a distance of $4^2+4^2-5^2=7$, but this is not the real distance which is $\sqrt{4^2+4^2} - 5= 0.65$ in this case. So the question is: what does the result of the circle function applied to a point not on the circle mean?

If we apply the formula with the point $(3,3)$ we get a negative number but what the heck does this number mean? The only thing I know is that if it’s $0$, it’s on the circle, if it’s negative it’s inside the circle and positive outside. But it doesn’t seem related the to the real distance so what does it represent?

1

There are 1 best solutions below

0
On

The point of the circle which is closest to $(4,4)$ is$$5\left(\frac4{\sqrt{4^2+4^2}},\frac4{\sqrt{4^2+4^2}}\right)=\left(\frac5{\sqrt2},\frac5{\sqrt2}\right)$$and therefore the distance from $(4,4)$ to the circle is$$\left\lVert(4,4)-\left(\frac5{\sqrt2},\frac5{\sqrt2}\right)\right\rVert=4\sqrt2-5.$$You can apply the same idea to $(3,3)$; you will get $5-3\sqrt2$.