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 $(0,0)$. The radius of the circle is $r=1$.The centre $(h,k)$ of the circle lies on the normal to th

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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 $(0,0)$. The radius of the circle is $r=1$.The centre $(h,k)$ of the circle lies on the normal to the tangent at $(0,0)$.The slope of the tangent $x+3y=0$ is $m_t = -\frac{1}{3}$.The slope of the normal is $m_n = -\frac{1}{m_t} = 3$.The equation of the normal passing through $(0,0)$ is $y-0 = 3(x-0)$,which simplifies to $y=3x$.Thus,the centre is of the form $(x, 3x)$.The distance from the centre $(x, 3x)$ to the point $(0,0)$ must be equal to the radius $r=1$.$\sqrt{(x-0)^2 + (3x-0)^2} = 1$$\sqrt{x^2 + 9x^2} = 1$$\sqrt{10x^2} = 1$$|x|\sqrt{10} = 1 \Rightarrow x = \pm \frac{1}{\sqrt{10}}$.If $x = \frac{1}{\sqrt{10}}$,then $y = 3x = \frac{3}{\sqrt{10}}$. The centre is $\left(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}\right)$.If $x = -\frac{1}{\sqrt{10}}$,then $y = 3x = -\frac{3}{\sqrt{10}}$. The centre is $\left(-\frac{1}{\sqrt{10}}, -\frac{3}{\sqrt{10}}\right)$.Comparing with the given options,the correct centre is $\left(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}\right)$.

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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