Question

The product of slope of tangents from point $(0,1)$ to the circle $x^2+y^2-2x+4y=0$ is

• A. $-1$
• B. $1$
• C. $-2$
• D. $2$

Answer 1

The correct option is A $-1$

Given equation of circle is
$x^2+y^2-2x+4y=0$
$\Rightarrow (x-1{)}^2+(y+2{)}^2=5$
$C\equiv (1,-2)$ and $r=\sqrt{5}$

Angle between the tangents from point $(0,1)$ is $2{\tan }^{-1}\left(\frac{r}{\sqrt{S_1}}\right)$
$\sqrt{S_1}=\sqrt{(0-1{)}^2+(1+2{)}^2-5}$
$\Rightarrow \sqrt{S_1}=\sqrt{5}=r$
$\therefore$ Angle between the tangents from point $(0,1)$ is $2\times \frac{\pi }{4}=\frac{\pi }{2}$
Hence product of their slopes will be $-1.$