question_answer

Let a and b be respectively the degree and order of the differential equation of the family of circles touching the lines, \[{{y}^{2}}-{{x}^{2}}=0\] and lying in the 1st and quadrant, then

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.,      \[a=2,\,\,b=1\]