Differential equation of the family of circles touching the line `y=2` at `(0,2)` is (a) `( b ) (c) (d) x^(( e )2( f ))( g )+( h ) (i)(( j ) (k) y-2( l ))^(( m )2( n ))( o )+( p )(( q ) dy)/( r )(( s ) dx)( t ) (u)(( v ) (w) y-2( x ))=0( y )` (z) (aa) `( b b ) (cc) (dd) x^(( e e )2( f f ))( g g )+(( h h ) (ii) y-2( j j ))(( k k ) (ll)2-2x (mm)(( n n ) dx)/( o o )(( p p ) dy)( q q ) (rr)-y (ss))=0( t t )` (uu) (vv) `( w w ) (xx) (yy) x^(( z z )2( a a a ))( b b b )+( c c c ) (ddd)(( e e e ) (fff) y-2( g g g ))^(( h h h )2( i i i ))( j j j )+(( k k k ) (lll) (mmm)(( n n n ) dx)/( o o o )(( p p p ) dy)( q q q ) (rrr)+y-2( s s s ))(( t t t ) (uuu) y-2( v v v ))=0( w w w )` (xxx) (yyy) None of these

Options

A. `x^(2)+(y-2)^(2)+(dy)/(dx)(y-2)=0`
B. `x^(2)+(y-2)(2-2x(dx)/(dy)-y)=0`
C. `x^(2)+(y-2)^(2)+((dx)/(dy)+y-2)(y-2)=0`
D. None of these

Answer

D

Explanation

Equation of family of circles will be `x^(2)+(y-2)^(2)+lambda(y-2)=0`
Differentiating, we get `2x+2(y-2)(dy)/(dx)+lambda(dy)/(dx)=0`
Thus,the equation is `x^(2)+(y-2)^(2)-(y-2) (2x(dx)/(dy) + 2y-4)=0`