A) \[2\left( \sqrt{2}+1 \right)\]
B) \[3+2\sqrt{2}\]
C) \[\sqrt{2}+1\]
D) \[2\left( 3+2\sqrt{2} \right)\]
Correct Answer: D
Solution :
tangent to\[{{x}^{2}}+{{y}^{2}}=4\] \[y=mx\pm 2\sqrt{1+{{m}^{2}}}\] \[{{x}^{2}}=4y\] \[{{x}^{2}}=4mx+8\sqrt{1+{{m}^{2}}}\] \[{{x}^{2}}=4mx-8\sqrt{1+{{m}^{2}}}=0\]D = 0 \[16{{m}^{2}}+4.8\sqrt{1+{{m}^{2}}}=0\] \[{{m}^{2}}+2\sqrt{1+{{m}^{2}}}=0\]or\[{{m}^{2}}=\sqrt{1+{{m}^{2}}}\] \[{{m}^{4}}=4+4{{m}^{2}}\] \[{{m}^{4}}-4{{m}^{2}}-4=0\] \[{{m}^{2}}=\frac{4\pm \sqrt{16+16}}{2}\] \[=\frac{4\pm 4\sqrt{2}}{2}\] \[{{m}^{2}}=2+2\sqrt{2}\]You need to login to perform this action.
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