What is the easier way to find the circle given three points?Given three points&nbsp;(x1,y1),(x2,y2), and&nbsp;(x3,y3), ify2−y1x2−x1≠y3−y2x3−x2≠y1−y3x1−x3,then there will be a circle passing through them. The general form of the circle isx2+y2+&nbsp;dx&nbsp;+ey+f=0.By substituting&nbsp;x=xi&nbsp;and&nbsp;y=yi, there will be a system of equation in three variables, that is:(x1y11x2y21x3y31)(def)=(−(x12+y12)−(x22+y22)−(x32+y32)).As there are a lot of things going around, the solution is prone to errors. Maybe this solution also has an error.Is there a better way to solve for the equation of the circle?

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What is the easier way to find the circle given three points?Given three points&amp;nbsp;(x1,y1),(x2,y2), and&amp;nbsp;(x3,y3), ify2−y1x2−x1≠y3−y2x3−x2≠y1−y3x1−x3,then there will be a circle passing through them. The general form of the circle isx2+y2+&amp;nbsp;dx&amp;nbsp;+ey+f=0.By substituting&amp;nbsp;x=xi&amp;nbsp;and&amp;nbsp;y=yi, there will be a system of equation in three variables, that is:(x1y11x2y21x3y31)(def)=(−(x12+y12)−(x22+y22)−(x32+y32)).As there are a lot of things going around, the solution is prone to errors. Maybe this solution also has an error.Is there a better way to solve for the equation of the circle?

umkhululiueyj · Accepted Answer

Move the circle so that it passes through the origin (subtract a point from all three). The equation loses a coefficient,x2+y2+&amp;nbsp;dx&amp;nbsp;+ey=0which is easier to solve. Then reverse-translate.−d2=|x12+y12y1x22+y22y2|2|x1y1x2y2|-e2=x1x12+y12x2x22+y222x1y1x2y2are the coordinates of the center in the moved position.

Terzago66cl · Accepted Answer

I prefer a different approach.You know that the center of the circle is the intersection between any two of the three perpendicular bisectors.Therefore,&amp;nbsp;(x−x1)2+(y−y1)2=(x−x2)2+(y−y2)2&amp;nbsp;&amp;nbsp;&amp;nbsp;(1)and&amp;nbsp;(x−x1)2+(y−y1)2=(x−x3)2+(y−y3)2&amp;nbsp;&amp;nbsp;&amp;nbsp;(2)Let&amp;nbsp;f1(x)=(x−x1)2−(x−x2)2.≈:≈&amp;nbsp;1st degree polynomialLet&amp;nbsp;f2(x)=(x−x1)2−(x−x3)2.≈:≈&amp;nbsp;1st degree polynomialLet&amp;nbsp;g1(y)=(y−y2)2−(y−y1)2.≈:≈&amp;nbsp;1st degree polynomialLet&amp;nbsp;g2(y)=(y−y3)2−(y−y1)2.≈:≈&amp;nbsp;1st degree polynomialYou can express&amp;nbsp;g2(y)&amp;nbsp;as&amp;nbsp;Ag1(y)+B, where A,B are fixed real numbers.This implies that&amp;nbsp;f2(x)=g2(y)=Ag1(y)+B=Af1(x)+B&amp;nbsp;&amp;nbsp;&amp;nbsp;(3)Equation (3) above gives you an equation solely in x, that you can solve to determine the&amp;nbsp;(x=x0)&amp;nbsp;coordinate of the center of the circle. Once this is determined, you can compute&amp;nbsp;y0&amp;nbsp;via&amp;nbsp;g1(y0)=f1(x0).

What is the easier way to find the

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