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…
[Math] find the equation to the circle circumscribing the quadrilateral formed by the straight lines
analytic geometrycircles
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Best Answer
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.