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Differential Equations Ncert Solutions Maths class 12th Maths Class 12th Differential Equations

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

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

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