Let a variable line passing through the centre of the circle x² + y² - 16x - 4y = 0, meet the positi

Q: Wait, why do we reject the root $a=4$?

Great catch! In pure algebra, $a=4$ is a valid solution to the derivative equation. But look at the geometry. If $a=4$, then $b = \frac{2(4)}{4-8} = -2$. A negative intercept means the line hits the *negative* y-axis. The problem explicitly states the line meets the *positive* coordinate axes. Always let the physical constraints guide your algebraic choices!

Q: Why did we use the intercept form $\frac{x}{a} + \frac{y}{b} = 1$?

It is the most elegant tool for this job. When a problem talks about intercepts on axes, the intercept form is your best friend. It directly maps the variables $a$ and $b$ to the geometry of the problem, making the substitution of the center point $(8, 2)$ immediate and clean.

Q: Is there a faster way than calculus?

Yes, the AM-GM inequality is a beautiful shortcut here! You could rewrite $S = a + b$ as $(a-8) + (b-2) + 10$. Then apply AM-GM to $(a-8)$ and $\frac{16}{a-8}$. However, calculus is a robust, universal tool that works even when inequalities aren't obvious. Mastering both is the mark of a true JEE champion.

Question

Let a variable line passing through the centre of the circle $x^2 + y^2 - 16x - 4y = 0$, meet the positive co-ordinate axes at the point $A$ and $B$. Then the minimum value of $OA + OB$, where $O$ is the origin, is equal to

Options:

(A) 12

(B) 18

(C) 20

(D) 24

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Line and the Circle

Welcome, future engineer. Today, we are going to peel back the layers of a problem that beautifully marries coordinate geometry with the art of optimization. It is not just about finding a number; it is about understanding how a line, constrained by a circle, dances across the axes.

### Phase 1: Decoding the Circle

We begin with the equation $x^2 + y^2 - 16x - 4y = 0$. To understand the circle, we must find its heart—the center.

By comparing this to the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify the center $C$ at $(8, 2)$. Imagine this point $(8, 2)$ as a fixed anchor in the first quadrant. Our variable line is tethered to this anchor, pivoting around it while slicing through the coordinate axes.

### Phase 2: The Variable Line

When a line cuts the axes at $A(a, 0)$ and $B(0, b)$, the most powerful tool in our arsenal is the intercept form:

$$ \frac{x}{a} + \frac{y}{b} = 1 $$

This equation is elegant because it directly relates the intercepts $a$ and $b$ to the line's position. Since the line passes through the center $C(8, 2)$, we substitute these coordinates into our equation:

$$ \frac{8}{a} + \frac{2}{b} = 1 $$

This is the constraint that binds our variables together.

### Phase 3: The Constraint and the Objective

We want to minimize the sum of the distances $OA + OB$, which is simply $S = a + b$. To minimize this, we need $S$ in terms of a single variable.

From our constraint, we isolate $b$:

$$ \frac{2}{b} = 1 - \frac{8}{a} = \frac{a-8}{a} \Rightarrow b = \frac{2a}{a-8} $$

Before we proceed, pause and look at the denominator. For $b$ to be positive (as the line hits the positive axes), $a-8$ must be positive. Thus, we have a vital domain constraint: $a > 8$.

### Phase 4: The Calculus of Optimization

Now, we define our objective function:

$$ S(a) = a + \frac{2a}{a-8} $$

To find the minimum, we take the derivative with respect to $a$:

$$ S'(a) = 1 + \frac{d}{da} \left( \frac{2a}{a-8} \right) $$

Applying the quotient rule, we get:

$$ S'(a) = 1 - \frac{16}{(a-8)^2} $$

Setting $S'(a) = 0$ leads us to $(a-8)^2 = 16$, which yields $a-8 = \pm 4$. This gives us $a = 12$ or $a = 4$.

Recalling our constraint $a > 8$, we immediately reject $a = 4$. The only valid critical point is $a = 12$.

### The Final Verdict

With $a = 12$, we find:

$$ b = \frac{2(12)}{12-8} = \frac{24}{4} = 6 $$

The minimum sum $OA + OB$ is $12 + 6 = 18$. We have successfully navigated the constraints and optimized the function.

**The minimum value of the sum $OA + OB$ is 18.**

Animated Solution for Mathematics - Differentiation: Let a variable line passing through the centre of the circle $x^{2} + y^{2} - 16 x - 4 y = 0$ , meet the positive co-ordinate axes at the point $A$ and $B$ . Then the minimum value of $O A + O B$ , where $O$ is the origin, is equal to

Select Answer:

Visualized Solution

Center of the Circle

Equation of the Variable Line

Line Passes Through Center

Isolating Variable $b$

Domain Constraint for $a$

The Objective Function $S (a)$

Derivative of $S (a)$

Solving for Critical Points

Selecting the Valid Root

Final Calculation of $O A + O B$

The Sigma Insight: Maxima and Minima

Analyzing the Geometry of the Line and the Circle

Phase 1

Phase 2

Phase 3

Phase 4

The Final Verdict

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