Question

equation of a common tangent to the circle x^2+y^2 -6x=0 and the parabola y^2 = 4x is

Answer 1

the answer of this question is S1=0

Answer 2

Answer:

Equation of Common tangent to given curves are y  = x/√3 + √3.

Step-by-step explanation:

Given:

Equation of Circle, x² + y² - 6x = 0

Rewriting given equation in standard form using completing the square method we get,

x² - 6x + y² = 0

x² - 6x + 3² + y² = 0 + 3²

( x - 3 )² + y² = 3²

Equation of parabola, y² = 4x

So, Equation of tangent ,T : y  = mx+ a/m

We know equation of parabola is y² = 4ax

So, 4a = 4 ⇒ a = 1.

⇒ y = mx + 1/m

Put, this in equation of circle.

$x^2-6x+(mx+\frac{1}{m})^2=0$

$x^2-6x+(mx)^2+(\frac{1}{m})^2+2\times mx\times\frac{1}{m}=0$

$(1+m^2)x^2-4x+\frac{1}{m^2}=0$

Now, We know that D ( Discriminant ) = 0

⇒ b² - 4ac = 0

$(-4)^2-4(1+m^2)(\frac{1}{m^2})=0$

$16m^2-4(1+m^2)=0$

$4m^2-(1+m^2)=0$

$4m^2-1-m^2=0$

3m² = 1

m = 1/√3

Equation of Common Tangent, T : y = 1/√3 x + 1/(1/√3)

y = x/√3 + √3

Therefore, Equation of Common tangent to given curves are y  = x/√3 + √3.