# Thread: center of a circle

1. ## center of a circle

Hi,
i think my question formation was wrong. i will ask it in different way. I have an arc . and the end points of the arc are say (x1,y1) and (x2,y2). i have the radius of the circle( R). Now i need to find out the center of the circle(X,Y). I need the equation of finding the center of circle using above information and the centre is not at origin. Thanking you in advance.

2. Originally Posted by prabhakar157
Hi,
i think my question formation was wrong. i will ask it in different way. I have an arc . and the end points of the arc are say (x1,y1) and (x2,y2). i have the radius of the circle( R). Now i need to find out the center of the circle(X,Y). I need the equation of finding the center of circle using above information and the centre is not at origin. Thanking you in advance.
Hello,

I don't know if this way may be of any help:

Let $x_c$ the x-value of the centre of the circle and
let $y_c$ the y-value of the centre of the circle and
let $x_1 , y_1$ the coordinates of a point on the circle

Than the equation $\left( x_c - x_1 \right)^2+\left( y_c - y_1 \right)^2=R^2$ describes the circle completely.

Put in the values you know. You'll get 2 equations which you can solve for $x_c$ and $y_c$.

Don't be afraid: The 2 equations look very ugly at the beginning but all squares will vanish rapidly.

Greetings

3. Originally Posted by earboth
Hello,

I don't know if this way may be of any help:

Let $x_c$ the x-value of the centre of the circle and
let $y_c$ the y-value of the centre of the circle and
let $x_1 , y_1$ the coordinates of a point on the circle

Than the equation $\left( x_c - x_1 \right)^2+\left( y_c - y_1 \right)^2=R^2$ describes the circle completely.

Put in the values you know. You'll get 2 equations which you can solve for $x_c$ and $y_c$.

Don't be afraid: The 2 equations look very ugly at the beginning but all squares will vanish rapidly.

Greetings
And there should be two solution points.

RonL

4. ## Thanks for ur reply

Hi,
Thanks for ur reply. i got the answer which i was looking for. Thank you.