# Circle through three points.xls

### Description

Let the three given points be a, b, c. Use ax and ay to represent the x and y coordinates of a, and so on. The coordinates of the center p=(px,py) of the circle determined by a, b, and c are:

To find the equation of a circle through three points, we can use the general equation of a circle:

(x-a)^2 + (y-b)^2 = r^2

where (a,b) is the center of the circle, and r is the radius.

We can use the three points to form a system of equations, with (a,b) and r as the unknowns. The system will have three equations and three unknowns. Solving this system will give us the equation of the circle.

Let the three points be (x1,y1), (x2,y2), and (x3,y3). Then we can write:

(x1-a)^2 + (y1-b)^2 = r^2 (x2-a)^2 + (y2-b)^2 = r^2 (x3-a)^2 + (y3-b)^2 = r^2

Expanding the squares, we get:

x1^2 - 2ax1 + a^2 + y1^2 - 2by1 + b^2 = r^2 x2^2 - 2ax2 + a^2 + y2^2 - 2by2 + b^2 = r^2 x3^2 - 2ax3 + a^2 + y3^2 - 2by3 + b^2 = r^2

Subtracting the first equation from the second and third equations, we get:

x2^2 - x1^2 - 2a(x2-x1) + y2^2 - y1^2 - 2b(y2-y1) = 0 x3^2 - x1^2 - 2a(x3-x1) + y3^2 - y1^2 - 2b(y3-y1) = 0

Multiplying the first equation by (y3-y1) and the second equation by (y2-y1), and subtracting the resulting equations, we get:

2a(y2-y3) + 2b(x3-x2) + x2^2 - x3^2 + y2^2 - y3^2 = 0

This is a linear equation in a and b. Solving for a or b and substituting into one of the original equations for r^2, we can find r^2. Once we have a, b, and r, we can plug them into the general equation of a circle to get the equation of the circle.

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