Let a variable line passing through the centre of the circle , meet the positive co-ordinate axes at the points A and B. Then the minimum value of OA + OB, where O is the origin is equal to [2024]

Question

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 points A and B. Then the minimum value of OA + OB, where O is the origin is equal to [2024]

Smart Achievers · Accepted Answer

(1)

Given circle is $x^{2} + y^{2} - 16 x - 4 y = 0$

Centre is (– g, – f)

Now, 2g = 16 $\Rightarrow$ g = 8 and 2f = 4 $\Rightarrow$ f = 2

$∴$ Centre is (8, 2)

Equation of line $l_{1}$ is ... (i)

y – 2 = m(x – 8)

Equation (i) cuts the x-axis then y = 0

$\frac{- 2}{m} = x - 8 \Rightarrow x = \frac{8 m - 2}{m} = O A$

Equation (i) cuts the y-axis, then x = 0

y – 2 = – 8m $\Rightarrow$ y = 2 – 8m = OB

Let $l = O A + O B = \frac{8 m - 2}{m} + 2 - 8 m = 10 - 8 m - \frac{2}{m}$ ... (ii)

Differentiate (ii) w.r.t. m we get

$\frac{d l}{d m} = - 8 + \frac{2}{m^{2}}$ ... (iii)

$\frac{d l}{d m} = 0 \Rightarrow - 8 + \frac{2}{m^{2}} = 0 \Rightarrow m = \pm \frac{1}{2}$

Differentiate (iii), w.r.t. m, we get

$\frac{d^{2} l}{d m^{2}} = \frac{- 4}{m^{3}}$

Now, $\frac{d^{2} l}{d m^{2}}$ $|_{m = \frac{- 1}{2}} = \frac{- 4}{{(\frac{- 1}{2})}^{3}} = 32 a n d \frac{d^{2} l}{d m^{2}} |_{m = \frac{1}{2}}$ $= \frac{- 4}{{(\frac{1}{2})}^{3}} = - 32$

$∴ \frac{d^{2} l}{d m^{2}} > 0 at m = \frac{- 1}{2}$

Hence, minima occurs at $m = \frac{- 1}{2}$

So, the minimum value of $l$ is

$= 10 - 8 \times (- \frac{1}{2}) - 2 (- 2) = 10 + 4 + 4 = 18$ .

Solution

Q.
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 points A and B. Then the minimum value of OA + OB, where O is the origin is equal to [2024]

1 18

2 20

3 12

4 24

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