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How does one solve the equation of a circle through three given points?...

(81,45) (81,-45) (85,0)

What is the solution to this???

I'm at a loss

cemeg47564
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  • We have several examples of solutions to such a question already. See https://math.stackexchange.com/questions/1018949/equation-of-circle-through-three-given-points, https://math.stackexchange.com/questions/827072/finding-an-equation-of-circle-which-passes-through-three-points, or https://math.stackexchange.com/questions/1745145/finding-equation-of-a-circle-given-three-non-collinear-points -- take your pick. – David K Jan 06 '20 at 13:24

2 Answers2

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Assume the equation to be of the form $(x-x_0)^2 + (y-y_0)^2 = r^2$ for some unknowns $x_0$, $y_0$ and $z_0$.

Plug in those three points for $(x, y)$ and solve for the unknowns.

Aryaman Maithani
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  • I have told you the procedure, you should substitute your points for $x$ and $y$. For example, $(81-x_0)^2+(45-y_0)^2 = r^2$ and similarly others. The symmetry will give you $y_0 = 0$ easily as pointed out in the other comment. You should be able to carry out the computations. – Aryaman Maithani Jan 06 '20 at 13:20
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The general equation of a circle is $$(x-a)^2+(y-b)^2=r^2,$$ with $(a,b)$ being the center of the circle, and $r$ being the radius.

Note that the first 2 points are symmetric about the $x-$axis, telling you that the y coordinate of the center is zero. So you get 2 equations with 2 unknowns $$(81-a)^2+45^2=r^2$$ $$(85-a)^2+0^2=r^2$$ You can subtract one equation from the other to solve for $a$, and then $r$ should be simple from there.

Paul
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