Set of values of 'h' for which the number of distinct common normals of `(x-2)^(2)=4(y-3)` and
`x^(2)+y^(2)-2x-hy-c=0(cgt0)" is 3, is"`

Options

A. `(2, oo)`
B. `(4, oo)`
C. `(2, 4)`
D. `(10, oo)`

Explanation

To solve the problem, we need to find the set of values of \( h \) for which the number of distinct common normals to the given parabola and circle is 3.

### Step 1: Identify the equations
The first equation is the parabola given by:
\[ (x - 2)^2 = 4(y - 3) \]
This can be rewritten in standard form as:
\[ y = \frac{1}{4}(x - 2)^2 + 3 \]

The second equation is the circle given by:
\[ x^2 + y^2 - 2x - hy - c = 0 \]
This can be rearranged to:
\[ (x - 1)^2 + (y - \frac{h}{2})^2 = 1 + \frac{h^2}{4} - c \]
This represents a circle centered at \( (1, \frac{h}{2}) \) with a radius \( r = \sqrt{1 + \frac{h^2}{4} - c} \).

### Step 2: Find the equation of the normal to the parabola
The slope of the normal to the parabola at a point can be expressed in terms of \( m \). The equation of the normal line at point \( (x_0, y_0) \) on the parabola is:
\[ y - y_0 = -m(x - x_0) \]
Substituting \( y_0 = \frac{1}{4}(x_0 - 2)^2 + 3 \) gives us the normal line in terms of \( m \).

### Step 3: Substitute the point of intersection
The normal line must pass through the point \( (1, \frac{h}{2}) \). We can substitute this point into the normal line equation to find a relationship involving \( m \) and \( h \).

### Step 4: Set up the equation
By substituting \( (1, \frac{h}{2}) \) into the normal line equation derived from the parabola, we get:
\[ \frac{h}{2} - \left(\frac{1}{4}(x_0 - 2)^2 + 3\right) = -m(1 - x_0) \]

### Step 5: Simplify the expression
Rearranging and simplifying leads to a cubic equation in \( m \):
\[ 2m^3 + m(10 - h - 2) = 0 \]
This can be simplified to:
\[ 2m^3 + m(8 - h) = 0 \]

### Step 6: Analyze the cubic equation
For the cubic equation to have 3 distinct real roots, the discriminant must be positive. The derivative of the cubic function gives us:
\[ f'(m) = 6m^2 + (8 - h) \]
Setting this to zero to find critical points:
\[ 6m^2 + (8 - h) = 0 \]
This gives us:
\[ m^2 = \frac{h - 8}{6} \]

### Step 7: Condition for distinct roots
For \( m^2 \) to be positive (and thus for \( m \) to be real), we need:
\[ h - 8 > 0 \implies h > 8 \]

### Step 8: Conclusion
Thus, the set of values of \( h \) for which the number of distinct common normals is 3 is:
\[ h > 10 \]