A) \[3x-2y=0\]
B) \[3x+2y=0\]
C) \[2x-3y=0\]
D) \[2x+3y=0\]
Correct Answer: B
Solution :
Given, circle is \[{{x}^{2}}+{{y}^{2}}-4x+6y=0\] On comparing with \[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0,\]we get \[2g=-4,2f=6\] \[\Rightarrow \] \[g=-2,f=3\] \[\because \]Equation of normal at point\[(x,\text{ }y)\]is \[\frac{x-{{x}_{1}}}{{{x}_{1}}+g}=\frac{y-{{y}_{1}}}{{{y}_{1}}+f}\] \[\therefore \]Equation of normal at (0, 0) is \[\frac{x-0}{0-2}=\frac{y-0}{0+3}\] \[\Rightarrow \] \[\frac{x}{-2}=\frac{y}{3}\] \[\Rightarrow \]\[3x=-2y\] \[\Rightarrow \] \[3x+2y=0\]
You need to login to perform this action.
You will be redirected in
3 sec
