question_answer

If \[\operatorname{y}=2x\] is a chord of the circle \[{{\operatorname{x}}^{2}}+{{y}^{2}}= 10 x\], then the equation of the circle whose diameter is this chord, is-

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 \[{{\operatorname{x}}^{2}}\,+{{y}^{2}}\,-\,2x-4y\,\,=0\]