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The equation of the circle circumscribing the triangle formed by the lines $y = 0, y = x$ and $2x + 3y = 10$ is?

I know this can be done by solving two equations at a time and finding the vertex. Then forming 3 different equations to solve for the centre (abscissa, ordinate) and the radius( which are 3 variables).

I would like to know if there's a shorter method to find the equation of the circle for any 3 random lines given. (It's easier in this particular case as one of the coordinates is $(0,0)$ which makes $c=0$)

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The equation of the circle is the following:

$$\operatorname{det}\left( \begin{bmatrix} \dfrac{1}{y} & 0 & 1 \\ \dfrac{2}{x-y} & 1 & -1 \\ \dfrac{13}{2x+3y-10} & 2 & 3\end{bmatrix} \right) = 0 \Longrightarrow \left(x-\dfrac{5}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2 = \dfrac{13}{2}$$

This comes from the generic formula that if you have three lines:

$$a_rx+b_ry+c_r=0, r=1,2,3$$

whose area of intersection forms a triangle, then the equation of the circle circumscribing the triangle is given by:

$$\operatorname{det}\left(\begin{bmatrix} \dfrac{a_1^2+b_1^2}{a_1x+b_1y+c_1} & a_1 & b_1 \\ \dfrac{a_2^2+b_2^2}{a_2x+b_2y+c_2} & a_2 & b_2 \\ \dfrac{a_3^2+b_3^2}{a_3x+b_3y+c_3} & a_3 & b_3\end{bmatrix}\right) = 0$$

Check out this post here for a proof.

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  • $\begingroup$ This is handy: it avoids having to compute the intersection points explicitly. $\endgroup$ – amd Jun 11 at 0:31
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Easy to see that $(0,0)$, $(2,2)$ and $(5,0)$ are the vertices of the triangle.

Let $M\left(\frac{5}{2},b\right)$ be the center of the circle.

Thus, $$\left(\frac{5}{2}\right)^2+b^2=\left(\frac{1}{2}\right)^2+(b-2)^2,$$ which gives $$b=-\frac{1}{2}$$ and $$M\left(\frac{5}{2},-\frac{1}{2}\right).$$ Now, for the radius of the circle we obtain: $$\sqrt{\frac{25}{4}+\frac{1}{4}}=\sqrt{\frac{13}{2}},$$ which gives the answer: $$\left(x-\frac{5}{2}\right)^2+\left(y+\frac{1}{2}\right)^2=\frac{13}{2}.$$

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Find the perpendicular bisectors of two of the sides of the triangle. They intersect at the center of the circle. And, of course the radius of the circle is the distance from that point to any vertex of the triangle. In the simple case you give here, the vertices of the triangle are (0, 0), (2, 2), and (5, 0). The midpoint of (0. 0) to (5, 0) is (5/2, 0) and the perpendicular bisector is x= 5/2. The midpoint of (0, 0) to (2, 2) is (1, 1) and the perpendicular bisector is y= -(x- 1)+ 1= 2- x. x= 5/2 and y= 2- x intersect at (5/2, 5/2). The distance from (5/2, 5/2) to (0, 0) is $5\sqrt{2}/2$ so the equation of the circumscribing circle is $(x- 5/2)^2+ (y- 5/2)^2= 25/2$.

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  • $\begingroup$ The centre of your circle is $(5/2,5/2)$, which can't be right, because it lies on the line $y=x$. $\endgroup$ – TonyK Jun 10 at 20:24
  • $\begingroup$ @TonyK The centre of a circumscribed circle to a triangle can lie on a side of the triangle... $\endgroup$ – Jean Marie Jun 10 at 21:33
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    $\begingroup$ @JeanMarie: Oh yes, you're right! Sorry. (But that can only happen for a right-angled triangle, which is not the case here.) $\endgroup$ – TonyK Jun 10 at 21:45
  • $\begingroup$ +1 for being the simplest method for this simple case. It will be even easier to understand if it comes with a diagram. $\endgroup$ – Mick Jun 11 at 9:18
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    $\begingroup$ $x=5/2$ and $y=2-x$ intersect at $(5/2,-1/2)$ because $y=2-5/2=-1/2$ and not at $(5/2, 5/2)$. The value of radius, is hence, also, wrong. $\endgroup$ – Tapi Jun 11 at 11:12

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