The loucs of the centre of the circle which cuts orthogonally the circle `x^(2)+y^(2)-20x+4=0` and which touches x=2 is

Options

A. `x^(2)=16y`
B. `x^(2)=16y+4`
C. `y^(2)=16x`
D. `y^(2)=16x+4`

Answer

To solve the problem step by step, we need to find the locus of the center of a circle that cuts orthogonally with the given circle and touches the line \( x = 2 \).

### Step 1: Rewrite the given circle equation
The equation of the given circle is:
\[ x^2 + y^2 - 20x + 4 = 0 \]
We can rewrite it in standard form by completing the square:
\[ (x^2 - 20x) + y^2 + 4 = 0 \]
\[ (x - 10)^2 - 100 + y^2 + 4 = 0 \]
\[ (x - 10)^2 + y^2 = 96 \]
This represents a circle with center \( (10, 0) \) and radius \( \sqrt{96} \).

### Step 2: General form of the second circle
Let the center of the second circle be \( (-g, -f) \) and its radius be \( r \). The equation of this circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

### Step 3: Condition for orthogonality
For two circles to cut orthogonally, the following condition must hold:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]
Substituting the values from our circles:
\[ 2(-g)(10) + 2(-f)(0) = c + 4 \]
This simplifies to:
\[ -20g = c + 4 \quad \text{(Equation 1)} \]

### Step 4: Condition for tangency to the line \( x = 2 \)
The distance from the center of the second circle to the line \( x = 2 \) must equal its radius:
\[ \text{Distance} = | -g - 2 | = r \]
The radius \( r \) can also be expressed as:
\[ r = \sqrt{g^2 + f^2 - c} \]
Thus, we have:
\[ | -g - 2 | = \sqrt{g^2 + f^2 - c} \quad \text{(Equation 2)} \]

### Step 5: Squaring both sides of Equation 2
Squaring both sides gives:
\[ (-g - 2)^2 = g^2 + f^2 - c \]
Expanding the left side:
\[ g^2 + 4g + 4 = g^2 + f^2 - c \]
This simplifies to:
\[ 4g + 4 = f^2 - c \quad \text{(Equation 3)} \]

### Step 6: Substitute Equation 1 into Equation 3
From Equation 1, we have \( c = -20g - 4 \). Substituting this into Equation 3:
\[ 4g + 4 = f^2 - (-20g - 4) \]
This simplifies to:
\[ 4g + 4 = f^2 + 20g + 4 \]
Subtracting \( 4 \) from both sides:
\[ 4g = f^2 + 20g \]
Rearranging gives:
\[ f^2 = -16g \quad \text{(Equation 4)} \]

### Step 7: Locus of the center
The locus of the center \((-g, -f)\) can be expressed as:
\[ f^2 = -16g \]
Substituting \( g = -x \) and \( f = -y \) gives:
\[ y^2 = 16x \]

### Final Answer
Thus, the locus of the center of the circle is:
\[ \boxed{y^2 = 16x} \]