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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