Question

Form differential equation of all circles touching both positive co-ordinate axes.

Accepted Answer

Equation of circle touching both positive coordinates axes is

${(x - a)}^{2} + {(y - a)}^{2} = a^{2} . . . . (i)$

Differentiating both sides w.r.t. $x$ , we get

$2 (x - a) + 2 (y - a) y^{'} = 0$

$\Rightarrow (x - a) + (y - a) y^{'} = 0$

$\Rightarrow (x + y y^{'}) - a (1 + y^{'}) = 0$

$\Rightarrow a = \frac{(x + y y^{'})}{(1 + y^{'})}$

Substituting in $(i)$ , we get

$\Rightarrow {\{x - \frac{(x + y y^{'})}{(1 + y^{'})}\}}^{2} + {\{y - \frac{(x + y y^{'})}{(1 + y^{'})}\}}^{2} = \frac{{(x + y y^{'})}^{2}}{{(1 + y^{'})}^{2}}$

$\Rightarrow {\{\frac{x (1 + y^{'}) - (x + y y^{'})}{(1 + y^{'})}\}}^{2} + {\{\frac{y (1 + y^{'}) - (x + y y^{'})}{(1 + y^{'})}\}}^{2} = \frac{{(x + y y^{'})}^{2}}{{(1 + y^{'})}^{2}}$

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

$\Rightarrow x^{2} {(1 + y^{'})}^{2} + {(x + y y^{'})}^{2} - 2 x (1 + y^{'}) (x + y y^{'}) + y^{2} {(1 + y^{'})}^{2} + {(x + y y^{'})}^{2} - 2 y (1 + y^{'}) (x + y y^{'}) = {(x + y y^{'})}^{2}$

$\Rightarrow (x^{2} + y^{2}) {(1 + y^{'})}^{2} + 2 {(x + y y^{'})}^{2} - 2 (x + y) (1 + y^{'}) (x + y y^{'}) = {(x + y y^{'})}^{2}$

$\Rightarrow (x^{2} + y^{2}) + \frac{2 {(x + y y^{'})}^{2}}{{(1 + y^{'})}^{2}} - \frac{2 (x + y) (1 + y^{'}) (x + y y^{'})}{{(1 + y^{'})}^{2}} = \frac{{(x + y y^{'})}^{2}}{{(1 + y^{'})}^{2}}$

$\Rightarrow (x^{2} + y^{2}) - \frac{2 (x + y) (x + y y^{'})}{(1 + y^{'})} + \frac{{(x + y y^{'})}^{2}}{{(1 + y^{'})}^{2}} = 0$

Embibe Experts Solutions for Chapter: Differential Equations, Exercise 1: Exercise-1

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Differential Equations, Exercise 1: Exercise-1

Questions from Embibe Experts Solutions for Chapter: Differential Equations, Exercise 1: Exercise-1 with Hints & Solutions

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