Question

If $x+2y-5$ is tangent to the circle $x^2+y^2=5$, then equation of normal at their point of contact is ?

Answer 1

Putting $x+2y=5$ in the equation of circle, we must get the point of contact.

$\Rightarrow (5-2y{)}^2+y^2=5$

$\Rightarrow 25+4y^2-20y+y^2=5$

$\Rightarrow 5y^2-20y+20=0$

$\Rightarrow y^2-4y+4=0$

$\Rightarrow (y-2{)}^2=0$

$\Rightarrow y=2$

$x=5-2y=5-4=1$

$\therefore$ Point of contact $\to (1,2)$

Normal passes through centre of circle.

$\Rightarrow$ Normalat $(1,2)$ must pass through $(0,0)$

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

$\Rightarrow y=2x$