The area (in sq. units) of the smaller of the two circles that touch the parabola, y² = 4x at the po

Q: Wait, why is the radius equal to $|k|$?

Because the circle touches the x-axis, the perpendicular distance from the center $(h, k)$ to the x-axis (which is the line $y=0$) is simply the vertical distance, which is $|k|$.

Q: Why does the center lie on the normal?

A circle touching a curve at a point must have its center on the line perpendicular to the tangent at that point. This line is the normal.

Q: Why do we reject one of the roots?

The quadratic equation $k^2 - 8k + 8 = 0$ yields two possible values for $k$, corresponding to two different circles. The problem asks for the 'smaller' circle, so we choose the smaller radius.

Question

The area (in sq. units) of the smaller of the two circles that touch the parabola, $y^2 = 4x$ at the point $(1, 2)$ and the x-axis is :-

Options:

(A) $4\pi(2-\sqrt{2})$

(B) $8\pi(3-2\sqrt{2})$

(C) $4\pi(3-\sqrt{2})$

(D) $8\pi(2-\sqrt{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem describes a circle that is tangent to the parabola $y^2 = 4x$ at the point $P(1, 2)$ and also tangent to the $x$-axis. Our objective is to determine the area of the smaller circle satisfying these conditions.

### The Shared Boundary

If a circle and a parabola touch at a single point, they must share a common tangent line at that point. We find the equation of the tangent to the parabola $y^2 = 4x$ at $P(1, 2)$ using the formula $yy_1 = 2a(x + x_1)$, where $a = 1$.

Substituting the given values:
$$y(2) = 2(1)(x + 1)$$

This simplifies to the linear equation:
$$x - y + 1 = 0$$

### The Path to the Center

The center of the circle must lie on the normal line passing through $P(1, 2)$. Since the tangent has a slope of $1$, the normal must have a perpendicular slope of $-1$.

Using the point-slope form $y - y_1 = m(x - x_1)$:
$$y - 2 = -1(x - 1) \Rightarrow x + y = 3$$

Let the center of the circle be $(h, k)$. Since the center lies on the normal, we have $h + k = 3$, or $h = 3 - k$. Thus, the center is represented as $(3 - k, k)$.

### The Tangency Condition

Because the circle is tangent to the $x$-axis, the radius $r$ is equal to the absolute value of the $y$-coordinate of the center, so $r = |k|$. Furthermore, the distance from the center $(3 - k, k)$ to the point $P(1, 2)$ must equal the radius $r$.

Setting up the distance formula squared:
$$(3 - k - 1)^2 + (k - 2)^2 = r^2$$

Substituting $r^2 = k^2$:
$$(2 - k)^2 + (k - 2)^2 = k^2$$

### The Algebra of the Solution

Since $(2 - k)^2 = (k - 2)^2$, the equation simplifies to:
$$2(k - 2)^2 = k^2$$
$$2(k^2 - 4k + 4) = k^2$$
$$k^2 - 8k + 8 = 0$$

Applying the quadratic formula $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$$k = \frac{8 \pm \sqrt{64 - 32}}{2} = 4 \pm 2\sqrt{2}$$

### Final Calculation

We have two possible values for the radius $r = k$. To find the area of the smaller circle, we select the smaller radius:
$$r = 4 - 2\sqrt{2}$$

The area $A$ is given by $\pi r^2$:
$$A = \pi (4 - 2\sqrt{2})^2$$
$$A = \pi (16 - 16\sqrt{2} + 8)$$
$$A = \pi (24 - 16\sqrt{2})$$

Factoring out the constant, the final area is:
**$8\pi(3 - 2\sqrt{2})$ square units.**

Animated Solution for Mathematics - Circles: The area (in sq. units) of the smaller of the two circles that touch the parabola, $y^{2} = 4 x$ at the point $(1, 2)$ and the x-axis is :-

Select Answer:

Visualized Solution

Visualizing the Problem

The Common Tangent

Equation of Tangent at $P (1, 2)$

The Normal Line at $P (1, 2)$

Defining the Center $C (h, k)$

The X-axis Tangency Condition

Distance from Center to $P (1, 2)$

Simplifying the Equation

Solving for $k$

Finding the Roots

Choosing the Smaller Circle

Final Area Calculation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Shared Boundary

The Path to the Center

The Tangency Condition

The Algebra of the Solution

Final Calculation

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area (in sq. units) of the smaller of the two circles that touch the parabola, y2=4x at the point (1,2) and the x-axis is :-