HSC Arts 12th Board ExamMaharashtra State Board
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Form the Differential Equation of the Family of Circles in the Second Quadrant and Touching the Coordinate Axes. - HSC Arts 12th Board Exam - Mathematics and Statistics

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Question

Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.

Solution

We know that (xa)2+(yb)2=r2 represents a circle with centre (ab) and radius r.​

Since the circle lies in the 2nd quadrant, and touches the coordinate axes, thus a < 0, b > 0 and |a| = |b| = r.

So, the equation becomes (x+a)2+(ya)2=a2  .....(1)

Differentiating this equation w.r.t. x, we get

`2(x+a)+2(y−a)dy/dx=0`

`⇒dy/dx=(−x+a)/(y−a)`

Putting `dy/dx=y',`  we get

`y'=(−x+a)/(y−a)`

`⇒yy'−ay'+x+a=0`

`⇒yy'+x=ay'−a`

`⇒a=(x+yy')/(y'−1)`

Substituting this value of a in (1), we get

`(x−(x+yy')/(y'−1))^2+(y−(x+yy')/(y'−1))^2=((x+yy')/(y'−1))^2`

`⇒(xy'−x−x−yy')^2+(yy'−y−x−yy')^2=(x+yy')^2`

`⇒[y'(x−y)−2x]^2+(x+y)^2=(x+yy')^2`

`⇒(y')^2(x^2−2xy+y^2)−4x^2y'+4xyy'+4x^2+x^2+2xy+y^2=x^2+2xyy'+y^2(y')^2`

`⇒(y')^2(x^2−2xy)+2xy'(−2x+y)+4x^2+2xy+y^2=0`

This is the required â€‹differential equation of the family of circles in the second quadrant and touching the coordinate axes.

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Solution Form the Differential Equation of the Family of Circles in the Second Quadrant and Touching the Coordinate Axes. Concept: General and Particular Solutions of a Differential Equation.
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