If the equation of one tangent to the circle | vedclass

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Question

(C) Let the equation of the other tangent be $y = mx$,which can be written as $mx - y = 0$.The center of the circle is $C(2, -1)$. The radius $r$ is t

Vedclass · Accepted Answer

$x - 3y = 0$

Vedclass · Answer

(C) Let the equation of the other tangent be $y = mx$,which can be written as $mx - y = 0$.The center of the circle is $C(2, -1)$. The radius $r$ is the perpendicular distance from $C$ to the tangent $3x + y = 0$.$r = \frac{|3(2) + 1(-1)|}{\sqrt{3^2 + 1^2}} = \frac{|6 - 1|}{\sqrt{10}} = \frac{5}{\sqrt{10}}$.Since the other tangent $mx - y = 0$ also touches the circle,the perpendicular distance from $C(2, -1)$ to this line must also be equal to $r$.$\frac{|m(2) - 1(-1)|}{\sqrt{m^2 + (-1)^2}} = \frac{5}{\sqrt{10}}$$\frac{|2m + 1|}{\sqrt{m^2 + 1}} = \frac{5}{\sqrt{10}}$Squaring both sides:$\frac{(2m + 1)^2}{m^2 + 1} = \frac{25}{10} = \frac{5}{2}$$2(4m^2 + 4m + 1) = 5(m^2 + 1)$$8m^2 + 8m + 2 = 5m^2 + 5$$3m^2 + 8m - 3 = 0$$(3m - 1)(m + 3) = 0$So,$m = \frac{1}{3}$ or $m = -3$.The given tangent is $3x + y = 0$,which has slope $m = -3$.Therefore,the other tangent has slope $m = \frac{1}{3}$.The equation is $y = \frac{1}{3}x$,which simplifies to $x - 3y = 0$.

If the equation of one tangent to the circle with center at $(2, -1)$ from the origin is $3x + y = 0$,then the equation of the other tangent through the origin is

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