If the line x+3y=0 is the tangent at (0,0) to | vedclass

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(D) The line $x+3y=0$ is tangent to the circle at $(0,0)$.Let the centre of the circle be $(h,k)$.Since the line passes through $(0,0)$ and is tangent

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

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(D) The line $x+3y=0$ is tangent to the circle at $(0,0)$.Let the centre of the circle be $(h,k)$.Since the line passes through $(0,0)$ and is tangent to the circle at $(0,0)$,the radius connecting the centre $(h,k)$ to the point of tangency $(0,0)$ must be perpendicular to the tangent line $x+3y=0$.The slope of the tangent line is $m_1 = -\frac{1}{3}$.The slope of the radius (normal) is $m_2 = \frac{k-0}{h-0} = \frac{k}{h}$.Since the radius is perpendicular to the tangent,$m_1 	imes m_2 = -1$,so $-\frac{1}{3} 	imes \frac{k}{h} = -1$,which gives $k = 3h$.The distance from the centre $(h,k)$ to the point $(0,0)$ is the radius,which is $1$.So,$\sqrt{h^2+k^2} = 1 \Rightarrow h^2+k^2 = 1$.Substituting $k=3h$,we get $h^2+(3h)^2 = 1$ $\Rightarrow 10h^2 = 1$ $\Rightarrow h = \pm \frac{1}{\sqrt{10}}$.If $h = \frac{1}{\sqrt{10}}$,then $k = \frac{3}{\sqrt{10}}$.If $h = -\frac{1}{\sqrt{10}}$,then $k = -\frac{3}{\sqrt{10}}$.Thus,the possible centres are $\left(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}ight)$ or $\left(-\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}ight)$.Comparing with the options,the correct centre is $\left(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}ight)$.

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

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