Question

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

• A. $ 6 (x^2 + y ^2) + 5 (3x - 2y) = 0 $
• B. $ x^ 2 + y ^2 - 2x - 3y + 5 = 0 $
• C. $ x^ 2 + y ^2 + 2x - 3y - 5 = 0 $
• D. $ 6 (x^2 + y^ 2) - 5 (3x + 2y) = 0 $

Answer 1

Answer: (D)

Given that, $x= 0, y = 0, 2x + 3y = 5$
On solving these equations, the required points
i.e., $A( 0, 0), (0 , 5/3 ), (5 /2 , 0 ) $.
Now, equation of circle is
$x^2 + y^2 + 2gx +2fy +c = 0 \quad ...(i)$
Eq. $(i)$ passes through $(0,0)$, we get $c = 0$.
Similarly, Eq. $(i)$ passes through $(0,5/3)$ and $(5/2,0)$, we get,
$ 2f = -5/2$ and $2g = -5/2$
∴ Required equation of circle is
$x^{2}+y^{2} - \frac{5}{2} x - \frac{5}{3}y = 0 $
or $6x^{2} +6y^{2} - 15 x -10y = 0$