Question

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

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

Answer 1

The correct option is B $6(x^2+y^2)-5(3x+2y)=0$

Given, triangle formed by the lines x = 0, y = 0, 2x + 3y = 5, so vertices of the triangle are (0, 0), (5/2, 0) and (0, 5/3).

Since circle is passing through (0, 0).

$\therefore$ Equation of circle will be $x^2+y^2+2gx+2fy=0$ ...(i)

Also, circle is passing through (5/2, 0) and (0, 5/3)
So, g = -5/4, f = -5/6.
Put the values of g and f in equation (i).
After solving, we get $6(x^2+y^2)-5(3x+2y)=0,$ which is the required equation of the circle.