A) \[4{{x}^{2}}+4{{y}^{2}}-30x-10y=15\]
B) \[4{{x}^{2}}+4{{y}^{2}}-30x-13y-25=0\]
C) \[4{{x}^{2}}+4{{y}^{2}}-17x-10y+25=0\]
D) None of the above
Correct Answer: D
Solution :
The required equation of circle is \[({{x}^{2}}+{{y}^{2}}+13x-3y)+\lambda \left( 11x+\frac{1}{2}y+\frac{25}{2} \right)=0\] ? (i) This circle passes through (1, 1). \[({{1}^{2}}+{{1}^{2}}+13-3)+\lambda \left( 11+\frac{1}{2}+\frac{25}{2} \right)=0\] \[\therefore \] \[12+\lambda (24)=0\] \[\Rightarrow \] \[\lambda =-\frac{1}{2}\] On putting this value of \[\lambda \] in Eq. (i), we get \[{{x}^{2}}+{{y}^{2}}+13x-3y-\frac{11}{2}x-\frac{1}{4}y-\frac{25}{4}=0\] \[\Rightarrow \]\[4{{x}^{2}}+4{{y}^{2}}+52x-12y-22x-y-25=0\] \[\Rightarrow \]\[4{{x}^{2}}+4{{y}^{2}}+30x-13y-25=0\]
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