To solve the problem of finding the area of the smaller of the two circles that touch the parabola \(y^2 = 4x\) at the point \((1, 2)\) and the x-axis, we can follow these steps:
### Step 1: Understand the Geometry
We have a parabola given by the equation \(y^2 = 4x\). The point of tangency is \((1, 2)\). We need to find the circles that touch both this parabola at the given point and the x-axis.
### Step 2: Find the Slope of the Tangent
To find the slope of the tangent to the parabola at the point \((1, 2)\), we differentiate the equation of the parabola:
\[
\frac{d}{dx}(y^2) = \frac{d}{dx}(4x) \implies 2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y}
\]
At the point \((1, 2)\):
\[
\frac{dy}{dx} = \frac{2}{2} = 1
\]
So, the slope of the tangent line at \((1, 2)\) is \(m_1 = 1\).
### Step 3: Find the Slope of the Normal
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[
m_2 = -\frac{1}{m_1} = -1
\]
### Step 4: Equation of the Normal Line
Using the point-slope form of the equation of a line, the equation of the normal at \((1, 2)\) is:
\[
y - 2 = -1(x - 1) \implies y = -x + 3
\]
### Step 5: Determine the Center of the Circle
Let the center of the circle be \((\alpha, \beta)\). Since the circle touches the x-axis, the distance from the center to the x-axis (which is \(\beta\)) is equal to the radius \(r\):
\[
\beta = r
\]
### Step 6: Find the Relationship Between \(\alpha\) and \(\beta\)
The slope of the line connecting the center \((\alpha, \beta)\) to the point of tangency \((1, 2)\) is given by:
\[
\frac{\beta - 2}{\alpha - 1} = -1
\]
This simplifies to:
\[
\beta - 2 = -(\alpha - 1) \implies \beta + \alpha = 3
\]
### Step 7: Substitute for \(\beta\)
From the previous step, we have:
\[
\beta = 3 - \alpha
\]
Substituting this into the equation for the radius:
\[
r = \beta = 3 - \alpha
\]
### Step 8: Equation of the Circle
The equation of the circle can be expressed as:
\[
(x - 1)^2 + (y - 2)^2 = r^2
\]
Substituting \(\beta\) gives:
\[
(x - 1)^2 + (y - (3 - \alpha))^2 = (3 - \alpha)^2
\]
### Step 9: Solve for \(\alpha\)
Expanding and rearranging gives us a quadratic equation in terms of \(\alpha\):
\[
\alpha^2 - 2\alpha - 7 = 0
\]
Using the quadratic formula:
\[
\alpha = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 28}}{2} = \frac{2 \pm 6}{2}
\]
This yields:
\[
\alpha = 4 \quad \text{or} \quad \alpha = -2
\]
Since \(\alpha\) must be positive, we take \(\alpha = 4\).
### Step 10: Find \(\beta\) and the Radius
Using \(\beta = 3 - \alpha\):
\[
\beta = 3 - 4 = -1 \quad \text{(not valid, so we use the other solution)}
\]
For the other solution, we can find the radius \(r\):
\[
\beta = 3 - (-2) = 5
\]
Thus, the radius \(r = 5\).
### Step 11: Calculate the Area of the Circle
The area \(A\) of the circle is given by:
\[
A = \pi r^2 = \pi (5)^2 = 25\pi
\]
### Final Answer
The area of the smaller circle that touches the parabola and the x-axis is:
\[
\boxed{25\pi}
\]
