i'm trying to determine the centrepoint of a circle which passes through two known points (x,y). also, the radius of the circle that passes through those two points is known.

with these given factors, how would i be able to calculate the center point of the circle?

  • $\begingroup$ Is that everything you know? That information does not uniquely describe one circle. $\endgroup$ – Dancrumb Oct 2 '17 at 12:52

You can also proceed with trigonometry. Call the distance between your two points $d$.

For any circle going through them, the following trigonometric equality holds: $$ r \sin \frac{\alpha}{2} = \frac{d}{2}$$ where $r$ is the radius, and $\alpha$ is the angle subtended by the arc though the two points, from the circle center. You know $r$ too, so $$\sin \frac{\alpha}{2} = \frac{d}{2r} $$ which allows to calculate $\alpha$.

By noting now that $r \cos \frac{\alpha}{2}$ is the distance from the mid-point of the chord to the center. Just erect a segment perpendicular to the chord (of course this can be done in two directions...), and of length $r \cos \frac{\alpha}{2}$, and you are on the centre.

The said distance (from the chord's midpoint to the centre) is of course also equal to $$\sqrt{r^2 - ( \frac{d}{2})^2} $$ maybe also useful.

  • $\begingroup$ okay so if i get it correctly: you "draw" a line between the two know points, then take the centrepoint of that line, and then use rcosα2rcos⁡α2 to calculate the centerpoint? $\endgroup$ – FrankK Oct 2 '17 at 13:09
  • $\begingroup$ also, how do i calculate α? @An aedonist $\endgroup$ – FrankK Oct 2 '17 at 13:28
  • $\begingroup$ @FrankK, to calculate $\alpha$ you have to invert the expression above containing $\sin \alpha /2$, $\frac{\alpha}{2} = asin \frac{d}{2r}$. You can also avoid calculating $\alpha$, have a look at the last equation, you can get the same result using Pythagora's Theorem and using $r$ and $d$ only. $\endgroup$ – An aedonist Oct 2 '17 at 13:50
  • $\begingroup$ @FrankK, in reference to your first comment, you draw a segment between your two points. You find its mid-point: through the midpoint, you draw a perpendicular line. The centre lies on the last drawn line, and the distance from the segment is given by the last expression in my post. $\endgroup$ – An aedonist Oct 2 '17 at 13:52
  • $\begingroup$ this is what i was looking for, thank you. $\endgroup$ – FrankK Oct 2 '17 at 14:00

Suppose $A$ and $B$ are the given points and $r$ the given radius.

The centre of the circle is on the perpendicular bisector $a$ of the segment $AB$

So I draw a circle with centre $A$ and radius $r$ which intersects $a$ in two points $C_1$ and $C_2$ which are the centres of the requested circles.

So there are two solutions, provided that $r> \dfrac{AB}{2}$.

If $r=\dfrac{AB}{2}$ there is only one solution: the circle having diameter $AB$

If $r< \dfrac{AB}{2}$ there are no solutions

Hope this can be useful


Suppose we have $A(1,2),\;B(3,4)$ and radius $r=2$

first write the perpendicular bisector $a$ that is the locus of the points $(x,y)$ such that $PA^2=PB^2$ $$(x-x_A)^2+(y-y_A)^2=(x-x_B)^2+(y-y_B)^2$$ plugging the data we get $a:x+y-5=0$

Now write the equation of the circle with centre $A$ and radius $r=2$


expand and get

$x^2+y^2-2 x-4 y+1=0$

Solve the system formed by the line $a$ and this last circle we have found

$ \left\{ \begin{array}{l} x^2+y^2-2x-4 y+1=0 \\ x+y-5=0 \\ \end{array} \right. $

Solutions are $C_1(1,\;4);\;C_2(3,\;2)$

so we can get the equations of the two circles

$$(x-1)^2+(y-4)^2=4;\;(x-3)^2+(y-2)^2=4$$ expanded they become $$x^2+y^2-2x-8y+13=0;\;x^2+y^2-6x-4y+9=0$$

enter image description here

  • $\begingroup$ hmm, i understand the concept, but i can't draw helper circles to determine the center point(S) of the circle(s) $\endgroup$ – FrankK Oct 2 '17 at 13:21
  • $\begingroup$ @FrankK what kind of data do you have. You didn't tell this information in the original post $\endgroup$ – Raffaele Oct 2 '17 at 13:39
  • $\begingroup$ given are: the (x,y) of two points on the circle, and the circle's radius. the required result is: the circle's centerpoint (x,y) $\endgroup$ – FrankK Oct 2 '17 at 13:43
  • $\begingroup$ @FrankK Look again at my answer: I have added the simplest way to solve the problem $\endgroup$ – Raffaele Oct 2 '17 at 14:07
  • $\begingroup$ I've already accepted the other answer, but this certainly enlightens me. thank you! $\endgroup$ – FrankK Oct 2 '17 at 14:16

Possibly a shorter way will be two first find the perpendicular bisector of the line segment joining the two points on the circle.

You know the length of the chord and radius. Using Pythagoras theorem you can find distance of center from chord.

Now use parametric equation of line to move this distance from centre of chord, along the perpendicular bisector.

You will essentially get two circles as the centre can lie on either side of the chord.


Straightforward method
Let center point be $(a,b)$. Then,

distance of center from one known point = radius
distance of center from second known point = radius

Use distance formula ($=\sqrt{(x_1-x_2)^2-(y_1-y_2)^2}$) in the above calculation. This yields two equations with two unknowns ($a$ and $b$). Square both sides to get:

$$(a-x_1)^2+(b-y_1)^2=r...E1$$ $$(a-x_2)^2+(b-y_2)^2=r...E2$$

Easier method: realize that the center of the circle will lie on the perpendicular bisector of the line joining these two points. This should give you an easier equation to work with.

Obviously, further realize that there would be two circles satisfying the above conditions.


enter image description here

Notice how the known points AB have become a chord to the circle.

  • $\begingroup$ im a bit confused, how does this give me the (x,y) of point C? $\endgroup$ – FrankK Oct 2 '17 at 13:13
  • $\begingroup$ @FrankK Which "this" are you exactly referring to? please specify clearly. $\endgroup$ – Gaurang Tandon Oct 2 '17 at 13:14
  • $\begingroup$ the straightforward method $\endgroup$ – FrankK Oct 2 '17 at 13:15
  • $\begingroup$ @FrankK Can you solve the two simultaneous equations I added above? $\endgroup$ – Gaurang Tandon Oct 2 '17 at 13:18
  • $\begingroup$ I'm afraid i don't remember how that works $\endgroup$ – FrankK Oct 2 '17 at 13:22

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