100. For the differential equation of the family of the circles in the first quadrant which touch the coordinate axes.

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Vishal Baghel | Contributor-Level 10

8 months ago

The equation of a circle in the first quadrant with centre (a, a) and radius (a) which touches the coordinate axes is:

${(x - a)}^{2} + {(y - a)}^{2} = a^{2} . . . . . . . . . . (1)$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l} 2 (x - a) + 2 (y - a) \frac{d y}{d x} = 0 \\ \Rightarrow (x - a) + (y - a) y^{'} = 0 \\ \Rightarrow x - a + y y^{'} - a y^{'} = 0 \\ \Rightarrow x + y y^{'} - a (1 + y^{'}) = 0 \\ \Rightarrow a = \frac{x + y y^{'}}{1 + y^{'}} \end{array}$

Substituting the value of a in equation (1), we get:

$\begin{array}{l} {[x - (\frac{x + y y^{'}}{1 + y^{'}})]}^{2} + {[y - (\frac{x + y y^{'}}{1 + y^{'}})]}^{2} = {(\frac{x + y y^{'}}{1 + y^{'}})}^{2} \\ \Rightarrow {[\frac{(x - a) y^{'}}{(1 + y^{'})}]}^{2} + {[\frac{y - x}{1 + y^{'}}]}^{2} = {[\frac{x + y y^{'}}{1 + y^{'}}]}^{2} \\ \Rightarrow {(x - y)}^{2} . y^{' 2} + {(x - y)}^{2} = (x + y y^{'})^{2} \\ \Rightarrow {(x - y)}^{2} [1 + (y^{'})^{2}] = (x + y y^{'})^{2} \end{array}$

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

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