Q. The equation of the circumcircle of the triangle formed by the lines $x = 0,\ y = 0,\ 2x + 3y = 5$ is
A) $6\left( x^{2} + y^{2} \right) + 5(3x - 2y) = 0$
B) $x^{2} + y^{2} - 2x - 3y + 5 = 0$
C) $x^{2} + y^{2} + 2x - 3y - 5 = 0$
D) $6\left( x^{2} + y^{2} \right) - 5(3x + 2y) = 0$
Explanation
(d)
On solving the given equations, we get$(0,\ 0),\ B(0,5/3)$ ,$C(5/2,\ 0)$.
Let equation of circle be
$x^{2} + y^{2} + 2gx + 2fy + c = 0$ …(i)
Eq. (i) passes through $A(0,\ 0)$, we get $c = 0$
Similarly, Eq. (i) passes through $B(0,5/3)$ and $C(5/2,\ 0)$, we get
$2f = - 5/3$ and $2g = - 5/2$
$\therefore$ Required equation of circle is
$x^{2} + y^{2} - \frac{5}{2}x - \frac{5}{3}y = 0$
$\Rightarrow \ \ 6x^{2} + 6y^{2} - 15x - 10y = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.