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

Options

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

Answer

To find the locus of the center of a circle that touches the line \(x - 2 = 0\) and cuts orthogonally the circle given by the equation \(x^2 + y^2 - 20x + 4 = 0\), we will follow these steps:

### Step 1: Identify the given line and circle
The line is given as \(x - 2 = 0\), which is a vertical line at \(x = 2\). The equation of the second circle can be rewritten in standard form.

### Step 2: Rewrite the circle equation
The equation of the circle is:
\[ x^2 + y^2 - 20x + 4 = 0 \]
We can complete the square for the \(x\) terms:
\[ (x^2 - 20x) + y^2 + 4 = 0 \implies (x - 10)^2 + y^2 = 100 - 4 \implies (x - 10)^2 + y^2 = 96 \]
Thus, the center of this circle is \((10, 0)\) and the radius is \(\sqrt{96} = 4\sqrt{6}\).

### Step 3: Identify the center of the circle that touches the line
Let the center of the circle we are looking for be \((g, f)\). Since this circle touches the line \(x - 2 = 0\), the distance from the center to the line must equal the radius \(r\):
\[ \text{Distance} = |g - 2| = r \]

### Step 4: Condition for orthogonality
For the circle to cut orthogonally with the given circle, the following condition must hold:
\[ 2(g_1 g_2 + f_1 f_2) = r_1^2 + r_2^2 \]
Here, \(g_1 = g\), \(f_1 = f\), \(g_2 = 10\), \(f_2 = 0\), \(r_1 = r\), and \(r_2 = 4\sqrt{6}\).

Thus, we have:
\[ 2(g \cdot 10 + f \cdot 0) = r^2 + (4\sqrt{6})^2 \]
This simplifies to:
\[ 20g = r^2 + 96 \]

### Step 5: Express \(r\) in terms of \(g\) and \(f\)
From the distance condition, we can express \(r\) as:
\[ r = |g - 2| \]
Squaring both sides gives:
\[ r^2 = (g - 2)^2 = g^2 - 4g + 4 \]

### Step 6: Substitute \(r^2\) into the orthogonality condition
Substituting \(r^2\) into the orthogonality condition:
\[ 20g = g^2 - 4g + 4 + 96 \]
This simplifies to:
\[ 20g = g^2 - 4g + 100 \]
Rearranging gives:
\[ g^2 - 24g + 100 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ g = \frac{24 \pm \sqrt{24^2 - 4 \cdot 1 \cdot 100}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 24^2 - 400 = 576 - 400 = 176 \]
Thus:
\[ g = \frac{24 \pm \sqrt{176}}{2} = \frac{24 \pm 4\sqrt{11}}{2} = 12 \pm 2\sqrt{11} \]

### Step 8: Find \(f\) in terms of \(g\)
From the earlier equation \(f^2 = -16g\), we can substitute \(g\) values to find \(f\).

### Step 9: Write the locus equation
The locus of the center \((g, f)\) can be expressed as:
\[ f^2 = -16g \]
This represents a parabola.

### Final Answer
The locus of the center of the circle is given by the equation:
\[ f^2 = -16g \]