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I have 2 points, e.g: (0,0) and (2,4) how do I know where is the center of the circle with its border pass over these 2 points

enter image description here

David K
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juan muñoz
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  • Don't think you can locate the center…three points are required to define a circle. Anymore information such as radius? – Li Chun Min May 19 '17 at 16:51
  • @LiChunMin That is not so in this case; see my solution below. – Cye Waldman May 19 '17 at 21:03
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    @CyeWaldman To be fair, you need to make some guesses about the intent of the figure in order to get the additional information you need. For example, it is sufficient to have two points plus the knowledge that you want the center to be on the $x$-axis. – David K May 19 '17 at 21:34
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    @LiChunMin I won't argue with that; in my own solution I relied on the fact that I knew the equation of the circle a priori. So, you are indeed correct; three pieces of information are needed. Even the other guy who did the solution required the foreknowledge that the center was on the $x$-axis. If you were given two random points in space, you could find many circles that pass thru them. To get a particular circle, you would need one more anchor. – Cye Waldman May 19 '17 at 21:46
  • Did you change the title? Still, with only two points, you can construct infinitely many circles with centers on the perpendicular bisector of the segment formed by $(0,0), (2,4)$… – Li Chun Min May 20 '17 at 00:41
  • Try draw it to see what it mean. Well, the radius is at least $\sqrt{5}$, in the case when that segment formed by those two given points are used as diameter. – Li Chun Min May 20 '17 at 00:43
  • If you assume center is on the x axis, then it is $(5,0)$ as pointed out in answer. – Li Chun Min May 20 '17 at 00:45

2 Answers2

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If $A$ and $B$ are two points on the circumference of a circle, then the center of the circle lies on the perpendicular bisector of the segment $\overline{AB}$.

When $A=(0,0)$ and $B= (2,4)$,

$\text{$\qquad$ Midpoint of segment $\overline{AB} = \left( \dfrac{0+2}{2}, \dfrac{0+4}{2} \right) = (1,2).$}$

$\text{$\qquad$ Slope of line $\overleftrightarrow{AB} = \dfrac{4-0}{2-0} = 2$}$

$\text{$\qquad$ Equation of perpendicular bisector: $y-2 = -\dfrac 12(x-1)$}$

According to your picture, the $y$-coordinate of the center of the circle is $0$.

So you get

\begin{align} y-2 &= -\dfrac 12(x-1) \\ 0-2 &= -\dfrac 12(x-1) \\ 4 &= x-1 \\ x &= 5 \end{align}

So the center is at $(5,0)$.

Steven Alexis Gregory
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Here is another kind of solution that I construct in the complex plane, though it could easily be done in Cartesian coordinates. But this is more succinct. It relies on the fact the I recognize that this circle is known in in polar coordinates to be $r=2R\cos\theta,\ \ \theta\in[0,\pi]$, where $R$ is the unknown radius in question.

Well, looking at the diagram, it's apparent that $r_B=2R\cos\theta_B=2R\cdot \Re(B)/r_B=|B|$, hence,

$$R=\frac{|B|^2}{2\Re(B)}$$

I have verified this result for randomly selected values $B=x+iy$. So the center of the circle is clearly at $x=R, y=0$. When $B=(2,4),\ \ R=5$.

Cye Waldman
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