A) \[{{\operatorname{x}}^{2}}+{{y}^{2}}+2x\,+4y=0\]
B) \[{{\operatorname{x}}^{2}}+{{y}^{2}}+2x-4y=0\]
C) \[{{\operatorname{x}}^{2}}+{{y}^{2}}-2x\,-4y=0\]
D) None of these
Correct Answer: C
Solution :
Here equation of the circle \[({{x}^{2}}+{{y}^{2}}-10x)\,\,+\,\lambda (y\,-2x)=\,\,0\] Now centre \[\operatorname{C} \left( 5 + \lambda , - \lambda /2 \right)\] lies on the chord again. \[\therefore \,\,\frac{-\lambda }{2}\,\,=\,\,2(5+\lambda )\] \[\therefore \text{ }\,\lambda =-\,4\] Hence \[{{\operatorname{x}}^{2}}\,+{{y}^{2}}\,\,=\,\,10x+4y\,-8x\] or \[{{\operatorname{x}}^{2}}\,+{{y}^{2}}\,-\,2x-4y\,\,=0\]You need to login to perform this action.
You will be redirected in
3 sec