A) \[(y-2)y{{'}^{2}}=25-{{(y-2)}^{2}}\]
B) \[{{(y-2)}^{2}}y{{'}^{2}}=25-{{(y-2)}^{2}}\]
C) \[{{(x-2)}^{2}}y{{'}^{2}}=25-{{(y-2)}^{2}}\]
D) \[(x-2)y{{'}^{2}}=25-{{(y-2)}^{2}}\]
Correct Answer: B
Solution :
[b] Circle \[{{(x-h)}^{2}}+{{(y-2)}^{2}}=25\] \[\Rightarrow 2(x-h)+2(y-2)y'=0\Rightarrow 2(y-2)y'\] \[=-2(x-h)\] \[\Rightarrow {{(y-2)}^{2}}y{{'}^{2}}={{(x-h)}^{2}}\Rightarrow {{(y-2)}^{2}}y{{'}^{2}}=25-{{(y-2)}^{2}}\]You need to login to perform this action.
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