A) \[8{{x}^{2}}+8{{y}^{2}}-24x+48y-13=0\]
B) \[16{{x}^{2}}+16{{y}^{2}}+24x-48y-13=0\]
C) \[16{{x}^{2}}+16{{y}^{2}}-24x+48y-13=0\]
D) \[8{{x}^{2}}+8{{y}^{2}}+24x-48y-13=0\]
Correct Answer: C
Solution :
The equation of given circle can be written as \[{{x}^{2}}+{{y}^{2}}-\frac{3}{2}x+3y+1=0\] whose centre is \[\left( \frac{3}{4},-\frac{3}{2} \right)\] and radius, \[r=\sqrt{\frac{9}{16}+\frac{9}{4}-1}\] \[=\sqrt{\frac{9+36-16}{16}}\] \[=\sqrt{\frac{29}{16}}\] \[\therefore \]Area of circle \[=\pi {{r}^{2}}\] \[=\pi \left( \frac{29}{16} \right)=\frac{29\pi }{16}\] \[\Rightarrow \]Area of required circle \[=2\times \frac{29\pi }{16}\] \[=\frac{29\pi }{8}\] Let R be the radius of required circle. \[\therefore \] \[{{R}^{2}}=\frac{29}{8}.\] Now, equation of circle is \[{{\left( x-\frac{3}{4} \right)}^{2}}+{{\left( y+\frac{3}{2} \right)}^{2}}=\frac{29}{8}.\] \[\Rightarrow \]\[{{x}^{2}}-\frac{3x}{2}+\frac{9}{16}+{{y}^{2}}+3y+\frac{9}{4}-\frac{29}{8}=0\] \[\Rightarrow \]\[16{{x}^{2}}+16{{y}^{2}}-24x+48y-13=0.\]
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