Htykuut

I need a Formula for this: we have two points on the circle. How can we find the center of circle?

For example $A(4.2,5.2)$, $B(5.2,6.3)$.

Two points do not uniquely determine a circle, so one cannot determine the center of a circle from only two points. It generally takes 3 points to uniquely determine a circle.

all we can say is that the centre of the circle lies on the perpendicular bisector of $AB$

HINT:

The distance of any point from the center = radius

Observe that we have two unknowns with one equation resulting in many solutions each corresponds to one circle.

Hint

Admitting that you have three points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$, the equation of the circle being $$(x-x_0)^2+(y-y_0)^2=r^2$$ you then have three equations for three unknowns $(x_0,y_0,r)$ $$F_1=(x_1-x_0)^2+(y_1-y_0)^2-r^2=0$$ $$F_2=(x_2-x_0)^2+(y_2-y_0)^2-r^2=0$$ $$F_3=(x_3-x_0)^2+(y_3-y_0)^2-r^2=0$$ So, expand $F_2-F_1$ and $F_3-F_1$; these give two linear equations in $x_0$ and $y_0$ which are easy to solve. Then replace in $F_1$ to get $r$ if you want it.

You can not find the center of the circle from two points alone. However, given $A$ and $B$ we know that the center of the circle lies on the perpendicular bisector of $AB$. Indeed the perpendicular bisector of $AB$ is exactly the locus of points equidistant to $A$ and $B$. Had you had three points, you would have three perpendicular bisectors, that just so happen to intersect in one point, which is the center of the circumscribed circle.

Letting $A=(x_1,y_1)$ and $B=(x_2,y_2)$, their center is at $(frac{x_1+x_2}{2},frac{y_1+y_2}{2})$. The slope of the perpendicular bisector is $frac{x_1-x_2}{y_2-y_1}$, so the equation of the line is:

$$y-frac{y_1+y_2}{2}=frac{x_1-x_2}{y_2-y_1}cdotleft(x-frac{x_1+x_2}{2}right)$$

Any coordinate pair satisfying this equation gives the center of some circle containing $A$ and $B$.

Basically, you can't. As others have noted, with two points you can find a whole line of possible center: the axis of the segment joining the two points. All the points on the axis are at the same distance from the two points.

Building the axis of the segment between two points $A,B$ is easy: draw the circumference centered at $A$ through $B$, draw the one through $A$ centered at $B$, intersect them and join the intersection, that line is the axis.

Finding its equation is a bit more tricky. Basically it's the same process done with calculations. If $A=(a_1,a_2)$ and $B=(b_1,b_2)$, then the circumferences are $(x-a_1)^2+(y-a_2)^2=r^2$ where $r^2=(b_1-a_1)^2+(b_2-a^2)^2$, and $(x-b_1)^2+(y-b_2)^2=r^2$. Expand the squares, set the LHS's equal and you should find two intersections. Say you get intersections $(x_1,y_1),(x_2,y_2)$. The line through these is the axis, and has equation $y-y_1=frac{y_2-y_1}{x_2-x_1}(x-x_1)$.

To find a single center, you need either another point (and thus you have $A,B,C$, find the axes of $AB$ and $AC$ or of any couple of segments you like best, and intersect them), or the circle drawn, so one axis passes through the center and thus if you take the segment of the axis lying within the circumference and draw the axis of that you find the center.

To be more precise, There are two possible centres for your circle, which are both equidistant from the line containing A and B and are equidistant from the perpendicular bisector of A and B.

Consider the points $A(x_1, y_1)$ and $B(x_2, y_2)$.

Now consider the loci radius $r$ around each point.

The equation of the locus around A will be $(x - x_1)^2 + (y - y_1)^2 = r^2$

The equation of the locus around B will be $(x - x_2)^2 + (y - y_2)^2 = r^2$

The point of intersection of these loci will be the points that are both a distance $r$ away from $A$ and $B$. This means that if you draw a circle radius $r$ around either point of intersection, the circle will contain both $A$ and $B$.

After doing the necessary mathematics, I figured out that the two centres of the circles are contained within the following quadratic formula in $y^2$ which return the resulting $y$-ordinates.

$(frac{(y_1 - y_2)^2}{(x_1 - x_2)^2} + 1)y^2 + frac{y_1 - y_2}{(x_1 - x_2)^2}((x_1^2 + y_1^2) - (x_2^2 + y_2^2) + 2x_1(x_1 - x_2))y + (frac{(x_2^2 + y_2^2) - (x_1^2 + y_1^2)}{2(x_1 - x_2)})^2 + x_1(frac{(x_1^2 + y_1^2) - (x_2^2 + y_2^2)}{(x_1 - x_2)}) + x_1^2 + y_1^2 - r^2 = 0$

Pretty equation, huh?

This is in the form $ay^2 + by + c = 0$

Which means to obtain the two $y$-ordinates, you're going to have to use the quadratic formula which is $y = frac{-bpmsqrt{{b^2 - 4ac}}}{2a}$

To then obtain the $x$-ordinate of the centre of the circle, simply substitute in the value of y into the equation obtained when subtracting the equation of the two loci.