• # 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

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.,      $a=2,\,\,b=1$