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प्रश्न
Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.
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उत्तर
We know that (x−a)2+(y−b)2=r2 represents a circle with centre (a, b) 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+(y−a)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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