Question

Let a variable line passing through the centre of the circle x2 + y2 – 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

• A. 12
• B. 18
• C. 20
• D. 24

Answer 1

Answer: (B)

18

Answer 2

The equation of the circle is:

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

Rewrite it in standard form by completing the square:

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

The center of the circle is $(8, 2)$.

Let the equation of the line passing through $(8, 2)$ be:

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

Find the intercepts. For the x -intercept, set $y = 0$:

$0 - 2 = m(x - 8)$ $x = \frac{-2}{m} + 8$

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

$y - 2 = m(0 - 8)$ $y = -8m + 2$

Calculate $OA + OB$. The distance $OA + OB$ is given by the sum of the intercepts:

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

Define $f(m) = \frac{-2}{m} + 8 - 8m + 2$. To find the minimum value, take the derivative $f'(m)$ and set it to zero:

$f'(m) = \frac{2}{m^2} - 8 = 0$

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

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

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

Thus, the minimum value of $OA + OB$ is:

18