Differential Equations — Class 12 Maths Solution

The equation of a circle in the first quadrant with centre $(a,a)$ and radius $$\left( a \right)$$ which touches the coordinate axes is:
${(x - a)^2} + {(y - a)^2} = {a^2}$

……(1)
Differentiating equation (1) with respect to $x$,

we get:
$2(x - a) + 2(y - a)\frac{{dy}}{{dx}} = 0$
$\Rightarrow$ $(x - a) + (y - a){y^\prime } = 0$
$\Rightarrow$ $x - a + y{y^\prime } - a{y^\prime } = 0$

$\Rightarrow$ $x + y{y^\prime } - a\left( {1 + {y^\prime }} \right) = 0$
$\Rightarrow$ $a = \frac{{x + y{y^\prime }}}{{1 + {y^\prime }}}$

Substituting the value of a in equation (1),

we get:
${\left[ {x - \left( {\frac{{x + y{y^\prime }}}{{1 + {y^\prime }}}} \right)} \right]^2} + {\left[ {y - \left( {\frac{{x + y{y^\prime }}}{{1 + {y^\prime }}}} \right)} \right]^2} = {\left( {\frac{{x + y{y^\prime }}}{{1 + {y^\prime }}}} \right)^2}$

$\Rightarrow$ ${\left[ {\frac{{(x - y){y^\prime }}}{{\left( {1 + {y^\prime }} \right)}}} \right]^2} + {\left[ {\frac{{y - x}}{{1 + {y^\prime }}}} \right]^2} = {\left[ {\frac{{x + y{y^\prime }}}{{1 + {y^\prime }}}} \right]^2}$

$\Rightarrow$ ${(x - y)^2} \cdot {y^{\prime 2}} + {(x - y)^2} = {\left( {x + y{y^\prime }} \right)^2}$

$\Rightarrow$ ${(x - y)^2}\left[ {1 + {{\left( {{y^\prime }} \right)}^2}} \right] = {\left( {x + y{y^\prime }} \right)^2}$

Hence, the required differential equation of the family of circles is ${(x - y)^2}\left[ {1 + {{\left( {{y^\prime }} \right)}^2}} \right] = {\left( {x + y{y^\prime }} \right)^2}$