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

The correct option is B $x+2y=1$

The given equation of circle is

$x^2+y^2=2x$

$\Rightarrow$ $x^2+y^2-2x=0$

$\Rightarrow$ $x^2+y^2-2x+0y+0=0$

Comparing it with general form of equation of circle

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

$\Rightarrow$ $2g=-2$

$\therefore$ $g=-1$

$\Rightarrow$ $2f=0$

$\therefore$ $f=0$

$\Rightarrow$ $c=0$

So center of circle $=C(-g,-f)=C(1,0)$

Equation of a given line is

$\Rightarrow$ $x+2y=3$ ------ ( 1 )

Slope of a line $\left(m\right)=\frac{-Coefficientofx}{Coefficientofy}=\frac{-1}{2}$

Now given that Normal is Parallel to this line, so

Slope of Normal $=m=\frac{-1}{2}$ (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 $=m=\frac{-1}{2}$.

So Equation of the Normal using $C(1,0)$ & $m=\frac{-1}{2}$

$y-y_1=m(x-x_1)$

$\Rightarrow$ $y-0=\frac{-1}{2}(x-1)$

$\Rightarrow$ $y=\frac{-1}{2}(x-1)$

$\Rightarrow$ $2y=-x+1$

$\Rightarrow$ $x+2y=1$