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

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Aicharya · Accepted Answer

Given line is $x+3 y=0$. $	herefore$ 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 . $\begin{array}{ll}	herefore & \frac{+g+3 f}{\sqrt{1^2+3^2}}=1 \ \Rightarrow & \frac{+g+3 f}{\sqrt{10}}=1\end{array}$ Taking option (d) i.e., centre $=\left(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}ight)$ $	herefore \quad \frac{+\frac{1}{\sqrt{10}}+3 	imes \frac{3}{\sqrt{10}}}{\sqrt{10}}=\frac{10}{10}=1$ (true)

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…

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