Solution:
assuming (p, 0) be the centre of the circle.
Therefore, it touches the y – axis at origin, its radius is p.
Since, the equation of the circle with centre (p, 0) and radius (p) is
![\[\begin{array}{*{35}{l}} \Rightarrow {{\left( x\text{ }-\text{ }p \right)}^{2}}~+\text{ }{{y}^{2}}~=\text{ }{{p}^{2}} \\ \Rightarrow {{x}^{2}}~+\text{ }{{p}^{2}}~-\text{ }2xp\text{ }+\text{ }{{y}^{2}}~=\text{ }{{p}^{2}} \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f35dee9ddfa0e141d63712c13ec4be36_l3.png "Rendered by QuickLaTeX.com")
Transposing p2 and – 2xp to RHS then it becomes – p2 and 2xp
![\[\begin{array}{*{35}{l}} \Rightarrow {{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }{{p}^{2}}~-\text{ }{{p}^{2}}~+\text{ }2px \\ \Rightarrow ~{{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }2px\text{ }\ldots \text{ }\left[ equation\text{ }\left( i \right) \right] \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fd68905b5591ea05accb0d9dbc2ccd5d_l3.png "Rendered by QuickLaTeX.com")
differentiating equation (i) both sides,
![\[\begin{array}{*{35}{l}} \Rightarrow 2x\text{ }+\text{ }2yy\text{ }=\text{ }2p \\ \Rightarrow ~x\text{ }+\text{ }yy\text{ }=\text{ }p \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-68a3b380427db07298d3b61108c69eff_l3.png "Rendered by QuickLaTeX.com")
substituting the value of ‘p’ in the equation, we get,
![\[\begin{array}{*{35}{l}} \Rightarrow {{x}^{2}}~+\text{ }{{y}^{2}}~=\text{ }2\left( x\text{ }+\text{ }yy \right)x \\ \Rightarrow ~2xyy\text{ }+\text{ }{{x}^{2}}~=\text{ }{{y}^{2}} \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-dbc9951f80df00111011c376469dff8b_l3.png "Rendered by QuickLaTeX.com")
Hence, 2xyy’ + x2 = y2 is the required differential equation.