The locus of the center of a circle which cuts orthogonally the parabola `y^2=4x` at (1,2) is a curve

Options

A. (3,4)
B. (4,3)
C. (5,3)
D. (2,4)

Explanation

To solve the problem of finding the locus of the center of a circle that cuts orthogonally to the parabola \( y^2 = 4x \) at the point \( (1, 2) \), we will follow these steps:

### Step 1: Find the Equation of the Tangent to the Parabola
The equation of the parabola is given by \( y^2 = 4x \). The formula for the equation of the tangent to the parabola at a point \( (x_1, y_1) \) is given by:

\[ yy_1 = 2(x + x_1) \]

Substituting the point \( (1, 2) \) into the equation:

\[ y \cdot 2 = 2(x + 1) \]

This simplifies to:

\[ 2y = 2x + 2 \]

Dividing through by 2 gives:

\[ y = x + 1 \]

### Step 2: Understand the Condition for Orthogonality
For a circle to cut the parabola orthogonally, the radius of the circle at the point of intersection must be perpendicular to the tangent of the parabola at that point. This means that the center of the circle lies on the tangent line we just found.

### Step 3: Set Up the Equation of the Circle
Let the center of the circle be \( (h, k) \). Since the center lies on the tangent line \( y = x + 1 \), we can express \( k \) in terms of \( h \):

\[ k = h + 1 \]

### Step 4: Find the Condition for Orthogonality
The general equation of a circle with center \( (h, k) \) and radius \( r \) is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

To find the condition for orthogonality, we can use the fact that the tangents to the circle and the parabola at the point of intersection should satisfy the orthogonality condition. The slope of the tangent to the parabola at \( (1, 2) \) is given by the derivative of the parabola at that point.

From the parabola \( y^2 = 4x \), we can differentiate implicitly:

\[ 2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y} \]

At the point \( (1, 2) \):

\[ \frac{dy}{dx} = \frac{2}{2} = 1 \]

The slope of the radius (from center \( (h, k) \) to \( (1, 2) \)) is:

\[ \frac{2 - k}{1 - h} \]

For orthogonality, the product of the slopes must equal \(-1\):

\[ 1 \cdot \frac{2 - k}{1 - h} = -1 \]

Substituting \( k = h + 1 \):

\[ \frac{2 - (h + 1)}{1 - h} = -1 \]

This simplifies to:

\[ \frac{1 - h}{1 - h} = -1 \]

### Step 5: Solve for the Locus
Cross-multiplying gives:

\[ 1 - h = - (1 - h) \]

This leads to:

\[ 1 - h = -1 + h \]

Combining like terms:

\[ 2h = 2 \implies h = 1 \]

Substituting back to find \( k \):

\[ k = 1 + 1 = 2 \]

### Conclusion
The locus of the center of the circle is a vertical line given by:

\[ h = 1 \]

Thus, the locus of the center of the circle that cuts orthogonally to the parabola at the point \( (1, 2) \) is:

\[ x = 1 \]