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I have the question "Find the equations of the following circles (in some cases more than one circle is possible).

A circle passes through the points $(1,4)$, $(7,5)$ and $(1,8)$"

I know that to find the equation of a circle I use $(X-a)^2 + (Y-b)^2 = r^2$,

However, I know you do this for one set of points I'm not sure what to do for three set of points and also the radius is not given.

The final answer should be $2X^2 + 2Y^2 -15X - 24Y + 77 = 0$.

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  • $\begingroup$ You do not have three sets of points. You have just three points. $\endgroup$ – KonKan Oct 10 '16 at 18:09
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Given two points on a circle, the center lies on the perpendicular bisector of the segment between them. Take two pairs of your points, find the equations of the bisectors and solve them simultaneously. One good choice is $(1,4)$ and $(1,8)$ because the bisector has a simple equation. The result will be the center, giving you $a$ and $b$. Then find the distance from that center to any one of your points, giving $r$.

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Since $A(1,4), B(7,5), C(1,8)$ lie on the circumference of the circle, the line segments $AB$, $BC$ are chords. Find the perpendicular bisectors of $AB$, $BC$. Call them $\epsilon_{AB},\epsilon_{BC}$. Determine the point of intersection of $\epsilon_{AB},\epsilon_{BC}$. This will give you the coordinates of the center. Then, compute its distance from any of the $A,B,C$. This will give you the radius.

P.S.: Of course a more straightforward method (but computationally more demanding), would be to simply substitute the coordinates of $A,B,C$ in the equation of the circle $(x-x_i)^2+(y-y_i)^2=r^2$, $i=A,B,C$ to obtain the simultaneous equations $$ (x-1)^2+(y-4)^2=r^2 \\ \\ (x-7)^2+(y-5)^2=r^2 \\ \\ (x-1)^2+(y-8)^2=r^2 \\ \\ $$ Solve them for $(x,y,r)$ and you are done. Note that, by elementary euclidean geometry considerations, the above system of simultaneous equations has always a unique solution, provided $A,B,C$ are not collinear.

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The triangle with vertices at $$A(1;4),\quad B(7;5),\quad C(1;8) $$ has side lengths $AB=\sqrt{37},\, AC=4,\, BC=3\sqrt{5}$ by the Pythagorean theorem and area $$[ABC]=\frac{1}{2}\left|1\cdot 5+7\cdot 8+1\cdot 4-4\cdot 7-5\cdot 1-8\cdot 1\right|=12 $$ by the shoelace formula. It follows that its circumradius is given by $$R=\frac{abc}{4[ABC]} = \frac{12\sqrt{185}}{48} = \frac{1}{4}\sqrt{185}. $$ The barycentric coordinates of the circumcenter are straightforward to compute through $a^2,b^2,c^2$, hence we know both the centre $$ O\left(3+\frac{3}{4};6\right)$$ and the radius of our circumcircle. Its equation readily follows.

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  • $\begingroup$ For the downvoter: please explain your downvote. What is wrong with this answer? $\endgroup$ – Jack D'Aurizio Oct 10 '16 at 19:31
  • $\begingroup$ I didn't down vote, but I'm guessing the down voter thought your solution was a bit more complex than it was supposed to. $\endgroup$ – Frank Oct 10 '16 at 19:50
  • $\begingroup$ @Frank: that is probably the reason, but I would like to underline that despite the fact that three non-trivial formulas (the shoelace formula, Euler's formula for the circumradius, the barycentric coordinates of the circumcenter) are involved, the resulting computations are very, very short, and they give us a way for finding the circumcenter of a triangle in $\mathbb{R}^n$ for $n\geq 3$, too. $\endgroup$ – Jack D'Aurizio Oct 10 '16 at 20:01
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    $\begingroup$ Moreover, since the supposed solution is quite trivial and probably already outlined in other answers, why not to shed some light on a non-trivial, but very effective, approach? I keep repeating these words: answers, on MSE, are not for the OP only. We are a community, not a homework-solving service for free. $\endgroup$ – Jack D'Aurizio Oct 10 '16 at 20:04
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The general form of any circle's equation is : x^2 + y^2 + 2gx + 2fy + c = 0

Substitute each of the given coordinates in this equation, and you will get 3 equations in f,g,c Solve for f,g,c and plug them in the above form to get your answer

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Translate the three points so that one becomes the origin. You get $(0,0), (6,1), (0,4)$.

The equation of a circle through the origin is

$$2(xx_c+yy_c)=x^2+y^2$$ and you get the $2\times2$ linear system

$$\begin{cases}12x_c+2y_c=37,\\8y_c=16,\end{cases}$$

giving $x_c=\dfrac{11}4,y_c=2,r^2=x_c^2+y_c^2=\dfrac{185}{16}$.

After restoring the origin, we get the claimed solution

$$\left(x-\frac{15}4\right)^2+\left(y-6\right)^2=\frac{185}{16}.$$

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In determinant form

$$ \begin{vmatrix} x^2+y^2 & x & y & 1 \\ x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ x_3^2+y_3^2 & x_3 & y_3 & 1 \end{vmatrix}=0$$

See also another answer of mine.

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