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

Explanation

To find the differential equation of the family of circles touching the line \( y = 2 \) at the point \( (0, 2) \), we can follow these steps:

### Step 1: Write the equation of the circle
The general equation of a circle with center at \( (0, k) \) and radius \( r \) can be expressed as:
\[ x^2 + (y - k)^2 = r^2 \]
Since the circle touches the line \( y = 2 \) at the point \( (0, 2) \), we can set \( k = 2 - r \) (the center must be below the line \( y = 2 \) by the radius \( r \)). Thus, the equation becomes:
\[ x^2 + (y - (2 - r))^2 = r^2 \]
This simplifies to:
\[ x^2 + (y - 2 + r)^2 = r^2 \]

### Step 2: Rearranging the equation
Expanding the equation:
\[ x^2 + (y - 2 + r)^2 = r^2 \]
\[ x^2 + (y - 2)^2 + 2r(y - 2) + r^2 = r^2 \]
Subtracting \( r^2 \) from both sides gives:
\[ x^2 + (y - 2)^2 + 2r(y - 2) = 0 \]

### Step 3: Introduce a parameter \( \lambda \)
Let \( r = \lambda \), then we have:
\[ x^2 + (y - 2)^2 + 2\lambda(y - 2) = 0 \]

### Step 4: Differentiate with respect to \( x \)
Differentiating the equation with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}((y - 2)^2) + \frac{d}{dx}(2\lambda(y - 2)) = 0 \]
This gives:
\[ 2x + 2(y - 2)\frac{dy}{dx} + 2\frac{d\lambda}{dx}(y - 2) + 2\lambda\frac{dy}{dx} = 0 \]

### Step 5: Rearranging the differentiated equation
Rearranging the differentiated equation:
\[ 2x + 2(y - 2)\frac{dy}{dx} + 2\lambda\frac{dy}{dx} = -2\frac{d\lambda}{dx}(y - 2) \]
Dividing through by 2:
\[ x + (y - 2)\frac{dy}{dx} + \lambda\frac{dy}{dx} = -\frac{d\lambda}{dx}(y - 2) \]

### Step 6: Eliminate \( \lambda \)
From the original equation, we can express \( \lambda \):
\[ \lambda = -\frac{x^2 + (y - 2)^2}{y - 2} \]
Substituting this back into the differentiated equation will yield the final differential equation.

### Final Differential Equation
After substituting and simplifying, the final differential equation will be:
\[ x^2 + (y - 2)^2 + 2(y - 2)\frac{dy}{dx} = 0 \]