Question

If the line $x+3y=0$ is the tangent at $(0,0)$ to circle of radius $1$, then the centre of one such circle is

• A. $(3,0)$
• B. $(\frac{-1}{\sqrt{10}},\frac{3}{\sqrt{10}})$
• C. $(\frac{3}{\sqrt{10}},\frac{-3}{\sqrt{10}})$
• D. $(\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}})$

Answer 1

The correct option is D $(\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}})$

Let $(h,k)$ be the centre of that circle.

Then equation of circle will be

${(x-h)}^2+{(y-k)}^2=1$

And $(h,k)$ will lie on normal an normal at origin will be perpendicular to tangent.

So, equation of tangent is $y=-\frac{x}{3}$

So, equation of normal $:y=3x$

$\therefore k=3h$

And origin lies on circle.

So, $h^2+k^2=1$

$\Rightarrow h^2+9h^2=1$

$\Rightarrow h=\pm \frac{1}{\sqrt{10}}$

So $K=\pm \frac{3}{\sqrt{10}}$

$\therefore$ Centres are $(\frac{1}{\sqrt{10}},\frac{3}{\sqrt{10}})$ or $(-\frac{1}{\sqrt{10}},-\frac{3}{\sqrt{10}})$