Centre of an arc

**ragsk** · February 18th 2006, 02:43 AM

Can ne 1 help me how to find the centre (Xc,Yc) for an arc having end points A(x1,y1)[say (2,5)] and B(x2,y2)[say (6,2)]?.........

I tried to calculate the centre imagining those end points of an arc as the end points of the diagonal of a square..........If we assume that the side of the square as 'a' then it becomes the radius 'r' for the arc which i found to be 5/sqrt(2) units with the given points (2,5) and (6,2)

The problem is, even if I knew the radius and the end points i'm unable to calculate the center C(Xc,Yc)...........
Was my calculation wrong or the approach was wrong?

And i also want to find the length 'l' of the arc and height 'h' of the curved surface................Please help me out.

I'm sending the figure as an attachment for clear picture of what i'm asking.

**earboth** · February 18th 2006, 05:23 AM

Originally Posted by ragsk

[SIZE=3][FONT=Palatino Linotype]Can ne 1 help me how to find the centre (Xc,Yc) for an arc having end points A(x1,y1)[say (2,5)] and B(x2,y2)[say (6,2)]?.........

Hello,

through 2 points you can draw many circles. All centres of these circles form a line perpendicular to AB.

So you need an additional value to get only 1 specific circle. Look at the attached drawing.

Greetings

EB

**topsquark** · February 18th 2006, 07:04 AM

As earboth says there are an infinity of circles matching your two points. A more algebraic approach will give you an idea of how to solve the problem in general and will show you why this happens.

I will assume the two points you gave: 1=(2,5) and 2=(6,2). The general equation for a circle is: $(x-x_c)^2+(y-y_c)^2=r^2$ where $(x_c,y_c)$ is the center of your circle and r is the radius. We have two points on that circle, 1 and 2, so we get two equations that $x_c$ , $y_c$ , and r must satisfy:

$(2-x_c)^2+(5-y_c)^2=r^2$
$(6-x_c)^2+(2-y_c)^2=r^2$

We need one more equation relating the three variables in order to solve this system in general. This is why we need more information to solve the problem.

However, if you are clever, it may appear we have such information: the center of the circle lies on the perpendicular bisector of the chord. To find this line, start by extending the chord to a line and find the equation of this line. Points 1 and 2 are on this line, so we can find it. Skipping the steps to the final result, this line is $y=- \frac{3}{4} x + 13/2$ .
The slope of the perpendicular to this line is $m'=4/3$ . Knowing the coordinates of the midpoint of the chord, M(4, 7/2), we can find the line that is the perpendicular bisector: $y= \frac{4}{3} x -11/6$ .
The center of the circle lies on this line so thus $y_c= \frac{4}{3} x_c -11/6$ .

This is, in fact, another relationship between the required variables, so our system is now:
$(2-x_c)^2+(5-y_c)^2=r^2$
$(6-x_c)^2+(2-y_c)^2=r^2$
$y_c= \frac{4}{3} x_c -11/6$

It may appear this is the solution to our difficulties. However there is a snag. Solve the last equation for $y_c$ and plug it into the top two equations. We find:
$(25/9)x^2-(200/9)x+(1825/36)=r^2$
$(25/9)x^2-(200/9)x+(1825/36)=r^2$

i.e. we get the same equation in both the two cases. This means our system is still indeterminate.

(The point of going through all of this is that you are going to need the perpendicular bisector of the chord to find h once you have the complete system of equations to solve your problem.)

-Dan

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