The equation of the circumcircle of the trian | vedclass

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(B) The triangle is formed by the lines $x = 0$ (y-axis),$y = 0$ (x-axis),and $2x + 3y = 5$.The vertices of the triangle are the intersection points o

Vedclass · Accepted Answer

$6(x^2 + y^2) - 5(3x + 2y) = 0$

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(B) The triangle is formed by the lines $x = 0$ (y-axis),$y = 0$ (x-axis),and $2x + 3y = 5$.The vertices of the triangle are the intersection points of these lines:$1$. Intersection of $x = 0$ and $y = 0$ is $(0, 0)$.$2$. Intersection of $y = 0$ and $2x + 3y = 5$ is $(5/2, 0)$.$3$. Intersection of $x = 0$ and $2x + 3y = 5$ is $(0, 5/3)$.Since the circle passes through the origin $(0, 0)$,its equation is of the form $x^2 + y^2 + 2gx + 2fy = 0$.Substituting the point $(5/2, 0)$ into the equation:$(5/2)^2 + 0 + 2g(5/2) + 2f(0) = 0 \implies 25/4 + 5g = 0 \implies g = -5/4$.Substituting the point $(0, 5/3)$ into the equation:$0 + (5/3)^2 + 2g(0) + 2f(5/3) = 0 \implies 25/9 + 10f/3 = 0 \implies 10f/3 = -25/9 \implies f = -5/6$.Substituting $g$ and $f$ back into the circle equation:$x^2 + y^2 + 2(-5/4)x + 2(-5/6)y = 0$$x^2 + y^2 - (5/2)x - (5/3)y = 0$Multiplying by $6$ to clear denominators:$6(x^2 + y^2) - 15x - 10y = 0$$6(x^2 + y^2) - 5(3x + 2y) = 0$.

The equation of the circumcircle of the triangle formed by the lines $x = 0$,$y = 0$,and $2x + 3y = 5$ is

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