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I have start point $(x_1,y_1)$ and an end point $(x_2,y_2)$ and radius of arc. How to calculate the co-ordinates of mid-poing of arc? The arc is the part of a circle.

enter image description here

Known Values

length of AD // that is radius
B(x,y)
C(x,y)

Needs to find

D(x,y)  // D is the mid-point of arc BC
Jorge
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coure2011
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    Well, if you know the coordinates of B and C, then you can find the midpoint M of the line segment BC, right? –  Jan 03 '12 at 19:57
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    go to $(B-C)/2$ and move perpendicular a distance of $\sqrt{r^2-|B-C|^2/4}$ to get to a center of a circle of radius $r$ through $A,B$. or solve $(a_1-h)^2+(a_2-k)^2=r^2,(b_1-h)^2+(b_2-k)^2=r^2$ for a center $(h,k)$, where $A=(a_1,a_2),B=(b_1,b_2)$ – yoyo Jan 03 '12 at 20:03
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    @coure2011 - There's no need to yell. You want help, right? Regardless, if you have the midpoint of the line segment BC, how can you use that to get the midpoint of the arc? –  Jan 03 '12 at 20:05

1 Answers1

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First assume the reduced coordinates $C=(0,y),B=(0,-y),D=(x,0)$ so that $A=(x-r,0)$. Then we solve

$$\|AC\|^2=r^2=(x-r)^2+y^2$$ which gives the two solutions

$$x=r\pm\sqrt{r^2-y^2},$$ of which you will prefer the $+$ sign.

enter image description here

Now to deal with the general position, notice that $y=\dfrac{\|BC\|}2$ gives you $x$, and you add to the midpoint of $BC$ $x$ times the unit vector perpendicular to $BC$, i.e.

$$D=\frac{B+C}2+\left(r+\sqrt{r^2-\|BC\|^2}\right)\dfrac{BC}{\|BC\|}\times 1_z.$$