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

Question

The equation of the circumcircle of the triangle formed by the lines x = 0,y = 0,2x + 3y = 5 is: (A)  6 (x2 + 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 (x2 + y 2) - 5 (3x + 2y) = 0

Tardigrade Admin · Accepted Answer

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 x2 + y2 + 2gx +2fy +c = 0 ...(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 â´ Required equation of circle is x2+y2 - (5/2) x - (5/3)y = 0 or 6x2 +6y2 - 15 x -10y = 0

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

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

$x^{2} + y^{2} - 2 x - 3 y + 5 = 0$

$x^{2} + y^{2} + 2 x - 3 y - 5 = 0$

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

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

6(x2+y2)+5(3x−2y)=0

x2+y2−2x−3y+5=0

x2+y2+2x−3y−5=0

6(x2+y2)−5(3x+2y)=0

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

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

$x^{2} + y^{2} - 2 x - 3 y + 5 = 0$

$x^{2} + y^{2} + 2 x - 3 y - 5 = 0$

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