A) a = 1, b = 2
B) a = 1, b = 1
C) a = 2, b = 1
D) a = 2, b = 2
Correct Answer: C
Solution :
Equation of circle is \[{{(x-0)}^{2}}+{{(y-k)}^{2}}={{\left( \frac{k}{\sqrt{2}} \right)}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+{{y}^{2}}-2ky+\frac{{{k}^{2}}}{2}=0\] \[\therefore \] Equation of tangent is \[2x+2yy'-2ky'=0\] \[k=\frac{x+yy'}{y'}\] From Eq. (i), we get \[{{x}^{2}}+{{y}^{2}}-2y\left( \frac{x+yy'}{y'} \right)+\frac{{{(x+yy')}^{2}}}{2{{(y')}^{2}}}=0\] \[\Rightarrow \] \[2{{(y')}^{2}}\,({{x}^{2}}+{{y}^{2}})-4yy'\,(x+yy')\] \[+{{(x+yy')}^{2}}=0\] \[\therefore \] Order =1, Degree = 2 i.e., \[a=2,\,\,b=1\]You need to login to perform this action.
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