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

• A

  $(3,0)$
• B

  $\left(\frac{-1}{\sqrt{10}}, \frac{3}{\sqrt{10}}\right)$
• C

  $\left(\frac{3}{\sqrt{10}}, \frac{-3}{\sqrt{10}}\right)$
• Correct answer
  D

  $\left(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}\right)$

Solution

Given line is $x+3 y=0$. $\therefore$ Slope of a line $=-\frac{1}{3}$ Let the centres of circle be $(+g,+f)$. We know that, the perpendicular drawn from the centre to the tangent is equal to radius.
Since perpendicular distance from $(g, f)$ to the line is 1 . Since perpendicular distance from $(g, f)$ to the line is 1 . $\begin{array}{ll}\therefore & \frac{+g+3 f}{\sqrt{1^2+3^2}}=1 \\ \Rightarrow & \frac{+g+3 f}{\sqrt{10}}=1\end{array}$ Taking option (d) i.e., centre $=\left(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}\right)$ $\therefore \quad \frac{+\frac{1}{\sqrt{10}}+3 \times \frac{3}{\sqrt{10}}}{\sqrt{10}}=\frac{10}{10}=1$ (true)

Hence, the correct answer is option: 'D'.