# Equation of a circle 3 points [duplicate]

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

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.
• 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 6 at 13:20
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.