Question

Fins the equation of normal to the circle x^2+y^2=2x which is parallel to the line x+2y=3

Answer 1

Given equation of circle is

x^2 + y^2 = 2x

x^2 + y^2 - 2x = 0

x^2 + y^2 - 2x + 0 y + 0 = 0

Comparing it with general form of equation of circle

x^2 + y^2 +2gx + 2fy + c = 0 , we get

2g = - 2       , 2f = 0    , c = 0

g = - 1          ,   f = 0    , c = 0

So center so circle is C(-g, -f) = C( 1, 0 )

Equation of a given line is

x+ 2y = 3    .................(i)

Slope of a line = m = - Coefficient of x/Coefficient of y

                        = -1/2

Now given that Normal is Parallel to this line , so

Slope of Normal = m = -1/2  (parallel lines have same slopes)

Now normal is a line that passes through the center of the circle and perpendicular to Tangent.

Now for required normal , we have point which is center of the circle C(1,0) and slope = m = -1/2.

So Equation of the Normal using C(1,0) & m=-1/2 in Point-Slope form is

  y-y1 = m ( x- x1)

 y - 0 = -1/2 (x - 1)

  y  = -1/2 (x - 1)

  2y = -1( x-1 )

  2y = -x + 1

x + 2y = 1

Answer 2

we have to find the equation of normal to the circle x² + y² = 2x which is parallel to the line x + 2y = 3

first of all, differentiate equation of circle with respect to x,

e.g., 2x + 2y.dy/dx = 2

⇒x + y.dy/dx = 1

⇒dy/dx = (1 - x)/y

we know, slope of normal in any curve y = f(x) is given by - dx/dy

so, slope of normal = -y/(1 - x)

A/C to question,

equation of normal to the circle is parallel to the line x + 2y = 3

so, slope of equation {x + 2y = 3} = slope of normal of curve { x² + y² = 2x}

⇒-1/2 = -y/(1 - x)

⇒ y/(1 - x) = 1/2

⇒ 2y = 1 - x

⇒ x + 2y = 1

Therefore the equation of normal to the curve x² + y² = 2x which is parallel to the line x + 2y = 3 is x + 2y = 1