Question

The equation of the normal to the circle \[{x^2} + {y^2} = 2x\] , which is parallel to the line \[x + 2y = 3\] is ,
A) \[x + 3y = 7\]
B) \[x + 2y = 1\]
C) \[x + 2y = 2\]
D) \[x + 2y = 5\]

Answer 1

Hint: In the problems involved a conic section and a straight line slope is obtained from the help of a straight line , and some additional information like a point is obtained with the help of conic section. Make use of both the information and get your required result.

Formula used:
\[y - {y_1} = m(x - {x_1})\]
\[{x^2} + {y^2} + 2gx + 2fy + c = 0\]
Slope of a line = \[m\] = $ - \dfrac{\text{coefficient of x}}{ \text{coefficient of y}}$

Complete step-by-step answer:
Given that , equation of the circle is \[{x^2} + {y^2} = 2x\]
\[{x^2} + {y^2} - 2x = 0\]
\[{x^2} + {y^2} - 2x + 0.y + 0 = 0\]
Now on comparing this equation with the general equation of the circle i.e \[{x^2} + {y^2} + 2gx + 2fy + c = 0\]
\[{x^2} + {y^2} + 2gx + 2fy + c = 0\]
\[{x^2} + {y^2} - 2x + 0.y + 0 = 0\]
\[2g = - 2;\;\;g = - 1\]
\[\ 2f = 0;\;\;f = 0 \\
  c = 0;\;\; \\
\ \]
So , we get the centre of the circle at \[C( - g, - f) = C(1,0)\]
We have also given the equation of the line
\[x + 2y = 3\]…………..…..(1)

Slope of a line = m = - coefficient of x / coefficient of y
                          m=-1/2 …..……..(2)
We have also given that the normal of the circle is parallel to this line , that means the slope of that line is also \[m = \dfrac{{ - 1}}{2}\]. Because parallel lines have the same slope.
We also know that the normal of the circle also passes through the centre of the circle and also it is perpendicular to the tangent .
Now for finding that equation of the normal line , we have a point that is the centre of the circle and a slope .
\[C( - g, - f) = C(1,0)\]
\[m = \dfrac{{ - 1}}{2}\]
So we can write the equation of the circle in point slope form
\[y - {y_1} = m(x - {x_1})\]
\[\
  y - 0 = - \dfrac{1}{2}(x - 1) \\
  y = - \dfrac{1}{2}(x - 1) \\
  2y = - 1(x - 1) \\
  x + 2y = 1 \\
\ \]
Hence the required equation of the line is \[x + 2y = 1\].

Option (B) is the correct option.

Note: We have given some additional formula:
Equation of the normal at point \[({x_1},{y_1})\]of the circle \[{x^2} + {y^2} + 2gx + 2fy + c = 0\] is

\[y - {y_1} = \left( {\dfrac{{{y_1} + f}}{{{x_1} + g}}} \right)(x - {x_1})\]
The equation of the normal on any point \[({x_1},{y_1})\]of the circle \[{x^2} + {y^2} = {a^2}\] is
\[\left( {\dfrac{y}{x} = \dfrac{{{y_1}}}{{{x_1}}}} \right)\]