Find equation of a circle based on 3 coordinates

• Dec 8th 2011, 11:53 PM
Lepzed
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!
• Dec 9th 2011, 12:32 AM
emakarov
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).
• Dec 9th 2011, 12:43 AM
Lepzed
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
• Dec 9th 2011, 12:43 AM
princeps
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 :

$\displaystyle (m,n,r)=(1,1,\sqrt 2)$
• Dec 9th 2011, 01:35 AM
emakarov
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.
• Dec 9th 2011, 08:22 AM
Lepzed
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?
• Dec 9th 2011, 10:14 AM
emakarov
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.
• Dec 9th 2011, 11:29 PM
Lepzed
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!