Question

The line , $\sf{ y = \sqrt{ 4 + 4 m^2 } }$ , such that m € R is tangent to thee circle -

1. x² + y² = 2

2. x² + y² = 4

3. x² + y² = 1

4. None of these .

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Answer 1

Answer:

your answer is here buddy.....

3.) x² + y² = 1

EXPLANATION-

y = √4+4m²

y² = 4 + 4m²

y² = 4(1+1m²)

y = 2√(1+m²).......(i)

we know that,

y = mx + a√1+m²

then substituting "a" value in (i).....

a = 2

how...

y = 2√(1+m²) & y= mx +a√1+m²

then, 2 = a

see the attachment for the diagram....

we have to find the expression for the the given tangent in respect to"x" & "y"....

given that m is an rational number,

let us take m as "x", because we need to find the relation of "x" & "y"...

so,

y² = 4+ 4x²

add x² on both side,

x² + y² = 4 + 4x² + x²

= 4 + 5x²

x² - 5x² + y² = 4

-4x² + y² = 4

x² + y² = 4/ -4

x² + y² = -1

we can write all the forms under the modules..

| x² + y² l = l -1 l

x² + y² = 1

hope this will help you, saby..

-------------THANK U------------

Answer 2

Hola mate

your answer is here buddy.....

y = √4+4m²

y² = 4 + 4m²

y² = 4(1+1m²)

y = 2√(1+m²).......(i)

we know that,

y = mx + a√1+m²

then substituting "a" value in (i).....

a = 2

how...

y = 2√(1+m²) & y= mx +a√1+m²

then, 2 = a

see the attachment for the diagram....

we have to find the expression for the the given tangent in respect to"x" & "y"....

given that m is an rational number,

let us take m as "x", because we need to find the relation of "x" & "y"...

so,

y² = 4+ 4x²

add x² on both side,

x² + y² = 4 + 4x² + x²

= 4 + 5x²

x² - 5x² + y² = 4

-4x² + y² = 4

x² + y² = 4/ -4

x² + y² = -1

we can write all the forms under the modules..

| x² + y² l = l -1 l

x² + y² = 1