A line is a common tangent to the circle (x - 3)² + y² = 9 and the parabola y² = 4x. If the two poin

Q: Wait, why can't I just use the condition $c^2 = r^2(1+m^2)$ for the circle?

That formula is a shortcut specifically for circles centered at the origin $(0,0)$. Since our circle is centered at $(3,0)$, that shortcut doesn't apply directly. We must use the general perpendicular distance formula from the center to the line.

Q: Why do we reject the negative slope $m = -\frac{1}{\sqrt{3}}$?

The problem specifies that the points of contact lie in the first quadrant. For the parabola $y^2 = 4x$, a tangent with a negative slope would touch in the fourth quadrant. Thus, we must choose the positive slope.

Q: Why did the quadratic equation for the circle's contact point result in a perfect square?

Because the line is tangent to the circle, it intersects at exactly one point. Algebraically, this means the quadratic equation must have a discriminant of zero, resulting in a repeated root.

Question

A line is a common tangent to the circle $(x - 3)^2 + y^2 = 9$ and the parabola $y^2 = 4x$. If the two points of contact $(a, b)$ and $(c, d)$ are distinct and lie in the first quadrant, then $2(a + c)$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at a circle centered at $(3,0)$ with a radius of $3$ and a parabola $y^2 = 4x$ opening to the right. A line is dancing between them, touching both curves perfectly.

We start by defining the tangent to the parabola. For any parabola $y^2 = 4ax$, the tangent with slope $m$ is given by $y = mx + \frac{a}{m}$.

With $a=1$, our equation simplifies to $y = mx + \frac{1}{m}$. Rearranging this, we obtain the general form:
$$m^2x - my + 1 = 0$$

### The Circle's Constraint

A line is tangent to a circle if and only if the perpendicular distance from the center of the circle to the line is exactly equal to the radius. Our circle is centered at $(3,0)$ with radius $r=3$.

Using the perpendicular distance formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$, we set the distance from $(3,0)$ to $m^2x - my + 1 = 0$ equal to $3$:
$$\frac{|3m^2 - m(0) + 1|}{\sqrt{(m^2)^2 + (-m)^2}} = 3$$

### Solving for the Slope

Squaring both sides to eliminate the absolute value and the square root, we get:
$$(3m^2 + 1)^2 = 9(m^4 + m^2)$$

Expanding the left side, we have:
$$9m^4 + 6m^2 + 1 = 9m^4 + 9m^2$$

The $9m^4$ terms cancel out, leaving us with $3m^2 = 1$, or $m^2 = \frac{1}{3}$. Since the points of contact are in the first quadrant, the slope must be positive, so $m = \frac{1}{\sqrt{3}}$.

### Finding the Points of Contact

With the slope $m = \frac{1}{\sqrt{3}}$ in hand, we find the contact point $(c, d)$ on the parabola using the formula $(\frac{a}{m^2}, \frac{2a}{m})$. Substituting $a=1$ and $m=\frac{1}{\sqrt{3}}$, we get:
$$c = \frac{1}{1/3} = 3, \quad d = \frac{2}{1/\sqrt{3}} = 2\sqrt{3}$$

Now for the circle's contact point $(a, b)$. We substitute the tangent line $y = \frac{1}{\sqrt{3}}x + \sqrt{3}$ into the circle's equation $(x - 3)^2 + y^2 = 9$:
$$(x - 3)^2 + \left(\frac{x+3}{\sqrt{3}}\right)^2 = 9$$

Simplifying this, we get $4x^2 - 12x + 9 = 0$, which is $(2x - 3)^2 = 0$. Thus, $x = \frac{3}{2}$, so $a = \frac{3}{2}$.

### The Final Celebration

We have found $a = \frac{3}{2}$ and $c = 3$. The problem asks for $2(a + c)$.

Plugging in our values:
$$2\left(\frac{3}{2} + 3\right) = 2\left(\frac{9}{2}\right) = 9$$

The elegance of the math reveals the **final answer is 9**. You have successfully navigated the geometry and the algebra.

Animated Solution for Mathematics - Conic Sections: A line is a common tangent to the circle $(x - 3)^{2} + y^{2} = 9$ and the parabola $y^{2} = 4 x$ . If the two points of contact $(a, b)$ and $(c, d)$ are distinct and lie in the first quadrant, then $2 (a + c)$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Curves

Defining the Tangent Equation

Condition for Circle Tangency

Setting up the Equation

Solving for the Slope

Selecting the Correct Slope

Point of Contact on Parabola

Point of Contact on Circle

Solving for coordinate $a$

Final Calculation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Circle's Constraint

Solving for the Slope

Finding the Points of Contact

The Final Celebration

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Enter Numerical Value:

Visualized Solution

Visualizing the Curves

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Solving for the Slope

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