JEE Mathematics 2024 Question with Solution

Given: The circle is

$$x^2 + y^2 - 16x - 4y = 0$$

and the variable line passes through its center.

Find: The minimum value of $OA + OB$ where the line meets the positive coordinate axes at A and B.

First rewrite the circle in standard form:

$$(x - 8)^2 + (y - 2)^2 = 68$$

So the center is $(8,2)$.

Let the line through $(8,2)$ be

$$y - 2 = m(x - 8)$$

For the x-intercept, put $y = 0$:

$$-2 = m(x - 8)$$ $$x = -\frac{2}{m} + 8$$

For the y-intercept, put $x = 0$:

$$y - 2 = -8m$$ $$y = -8m + 2$$

Hence,

$$OA + OB = \left|-\frac{2}{m} + 8\right| + \left|-8m + 2\right|$$

From the extracted solution, for the line to meet the positive axes and for the required minimum, we take

$$f(m) = -\frac{2}{m} + 8 - 8m + 2$$

Then

$$f'(m) = \frac{2}{m^2} - 8$$

Setting $f'(m)=0$,

$$\frac{2}{m^2} - 8 = 0$$ $$m^2 = \frac{1}{4}$$

So

$$m = -\frac{1}{2}$$

Substituting,

$$f\left(-\frac{1}{2}\right) = 18$$

Therefore, the minimum value of $OA + OB$ is $18$. Hence the correct option is B.

Note: The answer key says option C (20), but the solution concludes option B (18). the answer is taken as B.