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.You need to login to perform this action.
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