If y=3x is a tangent to a circle with centre | vedclass

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(A) The line $y-3x=0$ is a tangent to the circle. The radius $r$ is the perpendicular distance from the centre $(1,1)$ to the line $3x-y=0$.$r = \frac

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$3y=x$

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(A) The line $y-3x=0$ is a tangent to the circle. The radius $r$ is the perpendicular distance from the centre $(1,1)$ to the line $3x-y=0$.$r = \frac{|3(1) - 1(1)|}{\sqrt{3^2 + (-1)^2}} = \frac{|3-1|}{\sqrt{10}} = \frac{2}{\sqrt{10}}$.Let the other tangent passing through the origin $(0,0)$ be $y=mx$,or $mx-y=0$.The perpendicular distance from the centre $(1,1)$ to this line must also equal the radius $r$.$r = \frac{|m(1) - 1(1)|}{\sqrt{m^2 + (-1)^2}} = \frac{|m-1|}{\sqrt{m^2+1}}$.Equating the two expressions for $r$:$\frac{|m-1|}{\sqrt{m^2+1}} = \frac{2}{\sqrt{10}}$.Squaring both sides:$\frac{(m-1)^2}{m^2+1} = \frac{4}{10} = \frac{2}{5}$.$5(m^2 - 2m + 1) = 2(m^2 + 1)$.$5m^2 - 10m + 5 = 2m^2 + 2$.$3m^2 - 10m + 3 = 0$.$3m^2 - 9m - m + 3 = 0$.$3m(m-3) - 1(m-3) = 0$.$(3m-1)(m-3) = 0$.So,$m=3$ or $m=\frac{1}{3}$.Since $m=3$ corresponds to the given tangent $y=3x$,the other tangent is $y=\frac{1}{3}x$,which is $3y=x$.

If $y=3x$ is a tangent to a circle with centre $(1,1)$,then the other tangent drawn through $(0,0)$ to the circle is

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