Q. 53.9( 17 Votes )

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

Answer :

We know that the equation of a circle in the first quadrant with centre (a, a) and radius a which touches the coordinate axes is :

(x -a)^{2} + (y –a)^{2} = a^{2} -----------(1)

Now differentiating above equation w.r.t. x, we get,

2(x-a) + 2(y-a) = 0

⇒ (x – a) + (y – a)y’ = 0

⇒ x – a +yy’ – ay’ = 0

⇒ x + yy’ –a(1+y’) = 0

⇒ a =

Now, substituting the value of a in equation (1), we get,

⇒ (x - y)^{2}.y’^{2} + (x – y)^{2} = (x + yy’)^{2}

⇒ (x – y)^{2}[1 + (y’)^{2}] = (x + yy’)^{2}

Therefore, the required differential equation of the family of circles is

(x – y)^{2}[1 + (y’)^{2}] = (x + yy’)^{2}

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