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

(x -a)2 + (y –a)2 = a2 …………1

differentiating eq 1 with respect to x, we get,

\[\begin{array}{*{35}{l}}

2\left( x-a \right)\text{ }+\text{ }2\left( y-a \right)\text{ }dy/dx~=\text{ }0  \\

\Rightarrow ~\left( x\text{ }-\text{ }a \right)\text{ }+\text{ }\left( y\text{ }-\text{ }a \right)\text{ }y\text{ }=\text{ }0  \\

\end{array}\]

On multiplying we get

\[\begin{array}{*{35}{l}}

\Rightarrow ~x\text{ }-\text{ }a\text{ }+yy\text{ }-\text{ }ay\text{ }=\text{ }0  \\

\Rightarrow ~x\text{ }+\text{ }yy-\text{ }a\text{ }\left( 1+y \right)\text{ }=\text{ }0  \\

\end{array}\]

NCERT Solutions for Class 12 Maths Chapter 9 - Image 249

Therefore from above equation we have

NCERT Solutions for Class 12 Maths Chapter 9 - Image 250

NCERT Solutions for Class 12 Maths Chapter 9 - Image 251