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# Solution - Form the Differential Equation of the Family of Circles in the Second Quadrant and Touching the Coordinate Axes. - CBSE (Arts) Class 12 - Mathematics

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.​

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

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#### Reference Material

Solution for question: Form the Differential Equation of the Family of Circles in the Second Quadrant and Touching the Coordinate Axes. concept: null - General and Particular Solutions of a Differential Equation. For the courses CBSE (Arts), CBSE (Science), PUC Karnataka Science, CBSE (Commerce), HSC Science (General) , HSC Science (Computer Science), HSC Arts, HSC Science (Electronics), ISC (Commerce), ISC (Science), ISC (Arts)
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