Key Concepts and Formulas

• Equation of a Circle: The general form of a circle's equation is $x^2 + y^2 + 2gx + 2fy + c = 0$, with center $(-g, -f)$.
• Equation of a Line: The point-slope form of a line is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope. The intercept form is $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept.
• AM-GM Inequality: For non-negative numbers $a$ and $b$, $\frac{a+b}{2} \ge \sqrt{ab}$, with equality when $a = b$.

Step-by-Step Solution

Step 1: Find the Center of the Given Circle

The equation of the circle is $x^2 + y^2 - 16x - 4y = 0$. We compare this to the general form $x^2 + y^2 + 2gx + 2fy + c = 0$ to find the center.

$2g = -16 \implies g = -8$ $2f = -4 \implies f = -2$

The center of the circle is $(-g, -f) = (8, 2)$.

Why this step? The problem states the line passes through the center, so we need the center's coordinates to define the line.

Step 2: Write the Equation of the Variable Line

Since the line passes through $(8, 2)$, we can write its equation in point-slope form as $y - 2 = m(x - 8)$. However, since we are looking for the x and y intercepts, it is easier to work with the intercept form of the line. Let the intercepts on the x and y axes be $a$ and $b$ respectively. Then the equation of the line is given by $\frac{x}{a} + \frac{y}{b} = 1$.

Since the line passes through $(8, 2)$, we have $\frac{8}{a} + \frac{2}{b} = 1$.

Why this step? We need an equation for the line in terms of its intercepts $a$ and $b$, so we can relate them using the fact that the line passes through the circle's center.

Step 3: Express OA + OB in terms of a and b, and apply AM-GM

We want to minimize $OA + OB = a + b$. We are given that $a>0$ and $b>0$. From the equation in Step 2, we have $\frac{8}{a} + \frac{2}{b} = 1$. We wish to minimize $a+b$ subject to this constraint.

We can rewrite the constraint as $1 = \frac{8}{a} + \frac{2}{b} = \frac{a}{8} \cdot \frac{64}{a^2} + \frac{b}{2} \cdot \frac{4}{b^2}$.

Alternatively, consider $a+b = (a+b)(\frac{8}{a} + \frac{2}{b}) = 8 + \frac{8b}{a} + \frac{2a}{b} + 2 = 10 + \frac{8b}{a} + \frac{2a}{b}$.

Let $x = \frac{a}{8}$ and $y = \frac{b}{2}$. Then $\frac{8}{a} + \frac{2}{b} = 1$ becomes $\frac{1}{x} + \frac{1}{y} = 1$, so $x+y = xy$. We want to minimize $a+b = 8x+2y$.

Let's go back to $\frac{8}{a} + \frac{2}{b} = 1$. We want to minimize $a+b$. Using AM-GM on $\frac{8}{a}$ and $\frac{2}{b}$ is not directly helpful. Instead, we can write $a = OA$ and $b = OB$. We want to minimize $a+b$ subject to $\frac{8}{a} + \frac{2}{b} = 1$.

Using Cauchy-Schwarz inequality, $(a+b)(\frac{8}{a} + \frac{2}{b}) \ge (\sqrt{8} + \sqrt{2})^2 = (2\sqrt{2} + \sqrt{2})^2 = (3\sqrt{2})^2 = 18$. Since $\frac{8}{a} + \frac{2}{b} = 1$, we have $a+b \ge 18$. Equality holds when $\frac{a}{\sqrt{8}} = \frac{b}{\sqrt{2}}$, so $\frac{a}{2\sqrt{2}} = \frac{b}{\sqrt{2}}$, which implies $a = 2b$. Substituting into $\frac{8}{a} + \frac{2}{b} = 1$, we have $\frac{8}{2b} + \frac{2}{b} = 1$, so $\frac{4}{b} + \frac{2}{b} = 1$, which gives $\frac{6}{b} = 1$, so $b = 6$. Then $a = 2b = 12$. Therefore, the minimum value of $OA + OB = a + b = 12 + 6 = 18$.

Why this step? We needed to relate $a+b$ with $\frac{8}{a} + \frac{2}{b} = 1$ to minimize the sum. Cauchy-Schwarz is useful here.

Step 4: Another approach From $\frac{8}{a} + \frac{2}{b} = 1$, we have $\frac{2}{b} = 1 - \frac{8}{a} = \frac{a-8}{a}$, so $b = \frac{2a}{a-8}$. Since $b>0$, we must have $a>8$. We want to minimize $f(a) = a + b = a + \frac{2a}{a-8}$. Then $f'(a) = 1 + \frac{2(a-8) - 2a}{(a-8)^2} = 1 + \frac{-16}{(a-8)^2} = 0$. This gives $(a-8)^2 = 16$, so $a-8 = \pm 4$. Then $a = 12$ or $a = 4$. Since $a>8$, we have $a=12$. Then $b = \frac{2(12)}{12-8} = \frac{24}{4} = 6$. So $a+b = 12+6 = 18$. $f''(a) = \frac{32}{(a-8)^3}$. Since $a=12$, $f''(12) = \frac{32}{4^3} > 0$, so we have a minimum at $a=12$.

Why this step? This approach uses calculus to find the minimum. We needed to relate a and b, and find the critical points of the function.

Common Mistakes & Tips

• Sign Errors: Be careful with signs when finding the center of the circle and dealing with the slope.
• Constraint Overlook: Don't forget the constraint that the intercepts must be positive. This is crucial for determining the range of possible slopes.
• AM-GM Application: Correctly applying AM-GM or Cauchy-Schwarz is important for finding the minimum value. Make sure the conditions for equality can be satisfied.

Summary

We found the center of the circle and used it to relate the x and y intercepts $a$ and $b$ of a line passing through the center. We used Cauchy-Schwarz inequality to minimize $a+b$ subject to the constraint $\frac{8}{a} + \frac{2}{b} = 1$, obtaining a minimum value of 18. We also verified this using calculus.

Final Answer

The final answer is \boxed{18}, which corresponds to option (D).