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

The Correct Option is B

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