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How can I calculate center of a circle $x,y$? I have 2 points on the circumference of the circle and the angle between them.

The 2 points on the circle are $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$. The angle between them is $\theta$. I know all these values; how can I calculate the center from these values?

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what do you mean by the angle between the points? –  smanoos Jan 2 '12 at 18:30
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I guess he means the angle inscribed at centre by the line joining the two points. –  Nikhil Bellarykar Jan 2 '12 at 18:31
    
@NikhilBellarykar. Ahh ok. I see. He should have been more explicit. –  smanoos Jan 2 '12 at 18:37
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Note that all the answers give two solutions. This is correct-the center can be either side of the bisector line. –  Ross Millikan Jan 2 '12 at 19:13
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5 Answers 5

You could:

First, find the distance between, and the midpoint of, the two points.

Then draw the right triangle formed one of the given points, $p_1$, the midpoint, $m$, and the center, $c$, of the circle (this will be a right triangle since a perpendicular bisector of a chord on a circle passes through the center of the circle).

You know the length of the side $\overline{p_1m}$ of this triangle, since you know the distance between the given two points. You also know that the angle $\angle mcp_1$ is half the given angle.

Now, a bit of trig will allow you to find the side length $\overline{cm}$ of the triangle. Let's call that length $l$.

Next, find the equation of the line containing $\overline{cm}$ (it's slope will be the negative reciprocal of the slope of the line segment joining the two given points).

Let's say that equation is $y=m_0x+b$.

If $(c_1,c_2)$ are the coordinates of the center and $(m_1,m_2)$ are the coordinates of the midpoint, you'd know: $$ l = \sqrt{(c_1-m_1 )^2+(c_2-m_2)^2 } $$ and $$ c_2=m_0 c_1+b. $$ Finally, you'd solve the above equations for $c_1$ and $c_2$.

(There are probably slicker ways to do this.)


enter image description here

As Ross Millikan points out, there are two solutions (hence, there is another somewhat different diagram for the solution not represented by the above diagram)...

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what is b and m0 in second equation? –  coure2011 Jan 4 '12 at 10:08
    
@coure2011 The equation of the line $cm$ is $y= m_0 x+b$. $m_0$ is the slope of the line and $b$ is the $y$-intercept. $m_0$ is the negative reciprocal of the line segment $p_1p_2$. You can find the value of $b$ by substituting the values of the point $m$ into the equation. –  David Mitra Jan 4 '12 at 10:11
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It's quite easy using complex numbers. You can represent every point $(x,y)$ through the complex number $x+iy$ and vice versa. It's just easier to calculate with them this way.

Let $P_1 = 0$, then we get $S = \frac {P_2} 2$ as the starting point on the bisector. Multiply by $i$ to rotate by 90°, thus we need to go in the direction of $i\frac{S}{|S|}$. The distance is $d = \frac{|S|}{\tan \frac \theta 2}$, thus we get the center by

$$M = S + i\frac{S}{|S|}\cdot d = S(1 + \frac i {\tan \frac \theta 2})$$

Now for arbitrary points $P_1, P_2$, we get $$ M = \frac 1 2 (P_2-P_1) (1 + \frac i {\tan \frac \theta 2}) + P_1. $$

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What is itan? and M? –  coure2011 Jan 6 '12 at 5:39
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Hint:

Find the perpendicular bisector of $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$.

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I dont think this is enough. –  smanoos Jan 2 '12 at 18:28
    
Let $\theta$ be the angle between them. Then, starting from each point, make an angle of $(90-\theta)$ between the bisector. The point where the angle rays intersect the bisector is the centre. –  Nikhil Bellarykar Jan 2 '12 at 18:30
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Angle at centre is twice the angle between the given points

let centre be (x,y)

form equations for line segments joining centre to these 2 points respectively

you also have the ansgle between them, so solve to get (x,y)

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can you please explain it more? –  coure2011 Jan 6 '12 at 5:37
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The bisector of the line segment $\overline{P_1P_2}$ passes through the circle's center. Let $\alpha = \frac \theta 2$ and you get the distance $d$ to travel on the bisector by $$\tan \alpha = \frac{\overline{P_1P_2}}{d}.$$

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Should the numerator of the fraction be $\frac{1}{2}P_1P_2$, since the right triangle with $\alpha$ and $d$ is formed with the midpoint of $\overline{P_1P_2}$? –  Isaac Jan 3 '12 at 6:56
    
what this? how to find Cx and Cy center of circle? –  coure2011 Jan 6 '12 at 5:43
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