Question

If the line ax + y = c, touches both the curves x² + y² = 1 and y² - 4√(2) x , then |c| is equal to (A) 1/√2
(B) √2
(C) 1/2
(D) 2

Answer 1

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Answer 2

The value of $\left | c \right |$ for the given line is √2  .

Step-by-step explanation:

Given as :

The line equation is a x + y = c

The line touches the curves x² + y² = 1 and  y² = 4√2 x

Tangent to the  y² = 4√2 x is  y = m x + $\dfrac{\sqrt{2} }{m}$

And,

It is tangent to the circle x² + y² = 1

∴  $\left | \dfrac{\dfrac{\sqrt{2}}{m}}{\sqrt{1+m^{2}}} \right |$   =  1

Or, $\dfrac{\sqrt{2} }{m}$  = $\sqrt{1+m^{2} }$

Squaring both sides

[ $\dfrac{\sqrt{2} }{m}$ ]²  = [ $\sqrt{1+m^{2} }$ ]²

$\dfrac{2}{m^{2} }$   =  1 + m²

Or,  $m^{4}+m^{2}-2$  = 0

Or, $m^{4}$ + 2 m² - m² - 2 = 0

Or, m² ( m² + 2 ) - 1 ( m² +2 ) = 0

Or, ( m² +2 ) ( m² - 1 ) = 0

∴     m = 1 , m = - 1

So, Tangents are y = x + √2     ,    y = - x - √2

Now, Compare with  y = - a x + c

Therefore ,  a = $\pm$ 1   ,   c  =  $\pm$√2

So, The value of  $\left | c \right |$  = √2

Hence, The value of $\left | c \right |$ for the given line is √2  . Answer