find the equation to the circle circumscribing the quadrilateral formed by the straight lines

Question

find the equation to the circle circumscribing the quadrilateral formed by the straight lines

$$2x+3y=2$$

$$3x-2y=4$$

$$x+2y=3$$

$$2x-y=3$$

we can see that the first two and the last two are perpendicular...

Answer 1

First it would be wise to draw the lines, just to get a vizualization. Note that because none of the lines are parallel, they'll all intersect each other, so when you draw the lines you could determine the 4 vertices of the quadrilaterial. Now once you've done it find the coordinates of those points. Let their coordinates be: $(x_i,y_i); i = \overline{1,4}$

All this points are vertices, so they all line on the circumcircle. We know that every circle is defined by $3$ points and we know that the equation of the circle is:

$$(x-a)^2 + (y-b)^2 = r^2$$

Where $(a,b)$ are cooridnates of the circle's center. Now because all vertices lie on the circle just substitute and solve this system of 4 equation with 3 variables:

$$ \left\{\begin{aligned} &(x_1-a)^2 + (y_1-b)^2 = r^2\\ &(x_2-a)^2 + (y_2-b)^2 = r^2\\ &(x_3-a)^2 + (y_3-b)^2 = r^2\\ &(x_4-a)^2 + (y_4-b)^2 = r^2 \end{aligned} \right.$$

And the rest should be easy.

Answer 2

The diagonals of the quadrilateral will be the diameters of the circle, as they form right angles at the circumference. Therefore, after equating the equations of the lines and finding the end points of the diagonal, say $(x_1,y_1)$ , we can use the formula $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)=0$$