Question

Tangent to the circle x^2 +y^2=4 at any point on it i the first quadrant makes intercepts OA and OB on x and y axis respectively.O being the centre of the circle.find the minimum value of OA +OB.

Answer 1

0 must be the minimum value

Answer 2

Given :

Equation of circle :

${x}^{2} + {y}^{2} = 4$

Make intercepts OA and OB on x and y axis respectively.

To find :

Minimum value of OA + OB.

Solution :

For given equation :

${x}^{2} + {y}^{2} = 4$

(equation 1)

The equation of tangent for this equation :

$xx _{1} + yy _{1} = 4$

(equation 2)

where

$( x_{1} ,y _{1}) = (2 \cos \theta,2 \sin \theta)$

(equation 3)

When we have to know x- intercept,

we will put y = 0 in equation 2 then :

$xx_{1} = 4$

so

x-intercept

OA :

$x = \frac{4}{x_{1}} = \frac{4}{2 \cos \theta} = 2 \cosec \theta \:$

(equation 4)

When we have to know y- intercept,

we will put x = 0 in equation 2 then :

$yy_{1} = 4$

so

y-intercept

OB :

$y = \frac{4}{y_{1}} = \frac{4}{2 \sin \theta} = 2 \sec \theta \:$

(equation 5)

so, Let OA + OB = s

We have to minimise the sum of intercepts (s)

To get the minimum value

differentiating s with respect to θ.

(by equation 4 and 5)

$s = 2 \cosec \theta \: + 2 \sec \theta$

(equation 6)

$\frac{ds}{d \theta} = - 2 \cosec \theta \cot \theta \: + 2 \sec \theta \tan \theta$

Putting the above value =0 to find the value of θ

$- 2 \cosec \theta \cot \theta \: + 2 \sec \theta \tan \theta = 0 \\ \\ - 2( \frac{ \cos \theta}{ { \sin}^{2} \theta } ) +2 ( \frac{ \sin \theta}{ {\cos}^{2} \theta } ) = 0 \\ \\ { \sin}^{3} \theta = { \cos}^{3} \theta$

so

$\tan \theta = 1 \\ \theta = \frac{\pi}{4}$

Putting the value of θ in equation 6.

The minimum value of sum of intercepts:

s= OA + OB

$s = 2 \sec( \frac{\pi}{4} ) + 2 \cosec( \frac{\pi}{4} ) \\ \\ s = 2 \sqrt{2} + 2 \sqrt{2} \\ \\ \bf s = 4\sqrt{2}$

So the minimum value of sum of intercepts :

s = 4√2