Question

The differential equation of all the circles which touch both the coordinate axes in the first quadrant is

• A. $(x-y{)}^2(1+y^{\prime }{)}^2=(x+y{y}^{\prime }{)}^2$
• B. $(x-y{)}^2(1+\left(y^{\prime }{)}^2\right)=(x+y{y}^{\prime }{)}^2$
• C. $(x-y)(1+y^{\prime })=(x-y{y}^{\prime }{)}^2$
• D. $(x-y{)}^2(1-\left(y^{\prime }{)}^2\right)=(x+y{y}^{\prime }{)}^2$

Answer 1

The correct option is B $(x-y{)}^2(1+\left(y^{\prime }{)}^2\right)=(x+y{y}^{\prime }{)}^2$


Let $C(h,h)$ be the centre and $h$ be the radius of the circle.
Then, equation of the circle is
$(x-h{)}^2+(y-h{)}^2=h^2\cdots \left(1\right)$
Differentiating $\left(1\right)$ wrt $x,$ we get
$2(x-h)+2(y-h)y^{\prime }=0$
$\Rightarrow h=\frac{x+y{y}^{\prime }}{1+y^{\prime }}\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we get
$(x-y{)}^2[1+\left(y^{\prime }{)}^2\right]=(x+y{y}^{\prime }{)}^2$