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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

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Toolbox:
  • If the family of curves depends only on one parameter, then it is represented by an equation in the form $F(x,y,a) = 0$
  • Differentiating this with respect to x, we get an equation involving $y',x,y$ and a i.e $g(x,y,y') = 0$
  • The required differential equation is then obtained by eliminating a from the above equations.i.e $f(x,y,y') = 0$
Step 1:
The equation is of the form $(x-a)^2 + (y-a)^2 = a^2$-------(1)
On differentiating this with respect to $x$ we get,
$2(x-a)+2(y-a).y'=0$
dividing throughout by 2 we get,
$(x-a) + (y-a).y' = 0x+yy' = a(1+y')$
$a=\large\frac{(x+yy')}{(1+y')}$
Step 2:
Substituting this value of a in equ(1) we get
$[x -\large\frac{ (x+yy')}{(1+y')}]^2 +[y- \large\frac {(x+yy')}{(1+y')}]^2 = \big[\large\frac{x+yy'}{1+y'}\big]^2$
$[x(1+y') - x - yy']^2 + [y(1+y') - x - yy']^2 = [x+yy']^2$
On expanding we get,
$(xy' -yy')^2 + (y - x)^2 = (x+yy')^2$
$(x-y)^2(1 + y')^2 = (x+yy')^2$
This is the required solution.
answered Jul 30, 2013 by sreemathi.v
 

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