question_answer

The equation of the circumcircle of the triangle formed by the lines \[x=0,y=0,2x+3y=5\] is   [MP PET 2004]

A)            \[{{x}^{2}}+{{y}^{2}}+2x+3y-5=0\]                            

B)            \[6({{x}^{2}}+{{y}^{2}})-5\text{ }(3x+2y)=0\]

C)            \[{{x}^{2}}+{{y}^{2}}-2x-3y+5=0\]                              

D)            \[6({{x}^{2}}+{{y}^{2}})+5\text{ }(3x+2y)=0\]          

Correct Answer: B

Solution :

           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.