A) \[52{{r}^{2}}\]
B) \[20{{r}^{2}}\]
C) \[\frac{20}{9}{{r}^{2}}\]
D) \[\frac{52}{9}{{r}^{2}}\]
Correct Answer: A
Solution :
The line \[y=mx+c\] touches the circle\[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\] if and only if \[c=\pm r\sqrt{1+{{m}^{2}}}\] here we have line \[3x-2y=k\] \[\Rightarrow \] \[y=\frac{3}{2}x-\frac{1}{2}k\] and circle \[{{x}^{2}}+{{y}^{2}}=4{{r}^{2}}\] \[\therefore \] By condition \[c=\pm r\sqrt{1+{{m}^{2}}}\] we have \[-\frac{1}{2}k=\pm \,2r\sqrt{1+\frac{9}{4}}\] Squaring both sides \[\frac{1}{4}{{k}^{2}}=4{{r}^{2}}\left( \frac{13}{4} \right)\] \[{{k}^{2}}=52{{r}^{2}}\]
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