Find equation of a circle based on 3 coordinates

I've been given 3 coordinates: (0,0), (2,0) and (0,2) and I'm asked to draw the circle and then determine the equation of the circle: $\displaystyle (m - x)^2 + (n - y)^2 = r^2$ .

Now, since r=2 and the center point is (0,0), I'd say the equation would be:

$\displaystyle x^2 + y^2 = 4$, my book tells me it should be $\displaystyle x^2 + y^2 - 2x - 2y = 4$ (or something like that, don't have it with me). Could someone elaborate on this? I don't get where the -2x,-2y come from!

Re: Find equation of a circle based on 3 coordinates

Quote:

Originally Posted by

**Lepzed** I've been given 3 coordinates: (0,0), (2,0) and (0,2) and I'm asked to draw the circle and then determine the equation of the circle: $\displaystyle (m - x)^2 + (n - y)^2 = r^2$ .

I assume it's the circle that passes through these three points.

Quote:

Originally Posted by

**Lepzed** Now, since r=2 and the center point is (0,0)

No. The center lies on the perpendicular bisectors of the line segments (0,0) -- (0,2) and (0,0) -- (2,0), and the radius is the distance from the center to (0,0).

Re: Find equation of a circle based on 3 coordinates

No, one of the 3 given coordinates is the center, so in this case that would be (0,0), if I drew it correctly

Re: Find equation of a circle based on 3 coordinates

Quote:

Originally Posted by

**Lepzed** I've been given 3 coordinates: (0,0), (2,0) and (0,2) and I'm asked to draw the circle and then determine the equation of the circle: $\displaystyle (m - x)^2 + (n - y)^2 = r^2$ .

Now, since r=2 and the center point is (0,0), I'd say the equation would be:

$\displaystyle x^2 + y^2 = 4$, my book tells me it should be $\displaystyle x^2 + y^2 - 2x - 2y = 4$ (or something like that, don't have it with me). Could someone elaborate on this? I don't get where the -2x,-2y come from!

You should solve following system of equations :

$\displaystyle \begin{cases}m^2+n^2=r^2 \\(m-2)^2+n^2=r^2 \\m^2+(n-2)^2=r^2\end{cases}$

Solution is :

$\displaystyle (m,n,r)=(1,1,\sqrt 2)$

Re: Find equation of a circle based on 3 coordinates

Quote:

Originally Posted by

**Lepzed** No, one of the 3 given coordinates is the center, so in this case that would be (0,0), if I drew it correctly

If the circle equation is $\displaystyle x^2+y^2-ax-by+c=0$ for some $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$, then the circle center is $\displaystyle (a/2,b/2)$. So, your claim that one of the given coordinates is the center contradicts the supposed answer $\displaystyle x^2 + y^2 - 2x - 2y = \dots$. It also does not follow from the problem statement in the first sentence of the OP.

Re: Find equation of a circle based on 3 coordinates

Took a while for me to get home (and to my book). The three coordinates given are indeed (0, 0), (2, 0) and (0, 2). And indeed you were right, none of these points is supposed to be the center. But that makes me wonder, how can i possibly draw a circle on which all these points lie?

Re: Find equation of a circle based on 3 coordinates

Quote:

Originally Posted by

**Lepzed** But that makes me wonder, how can i possibly draw a circle on which all these points lie?

Every triangle has a circle that passes through all its three vertices. As said in post #4, the circle in question has center (1, 1) and radius $\displaystyle \sqrt{2}$. In fact, when the triangle has a right angle, the circumcircle's center is the middle of the hypotenuse.

Re: Find equation of a circle based on 3 coordinates

Ah, yeah, I was kind of stuck in a wrong train of thought there! Cheers!