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

ConceptGeneral and Particular Solutions of a Differential Equation

#### 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.â€‹

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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