The equations of the tangents to the circle x | vedclass

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(A) line passing through the point $(4,0)$ with slope $m$ is given by $y - 0 = m(x - 4)$,which simplifies to $mx - y - 4m = 0$.For this line to be a t

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$y = \pm \frac{1}{\sqrt{3}}(x-4)$

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(A) line passing through the point $(4,0)$ with slope $m$ is given by $y - 0 = m(x - 4)$,which simplifies to $mx - y - 4m = 0$.For this line to be a tangent to the circle $x^2 + y^2 = 4$ (with center $(0,0)$ and radius $r = 2$),the perpendicular distance from the center to the line must equal the radius.Using the distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$,we have $\frac{|m(0) - (0) - 4m|}{\sqrt{m^2 + (-1)^2}} = 2$.$| -4m | = 2\sqrt{m^2 + 1}$.Squaring both sides: $16m^2 = 4(m^2 + 1)$ $\Rightarrow 4m^2 = m^2 + 1$ $\Rightarrow 3m^2 = 1$ $\Rightarrow m = \pm \frac{1}{\sqrt{3}}$.Substituting $m$ back into the line equation: $y = \pm \frac{1}{\sqrt{3}}(x - 4)$.

The equations of the tangents to the circle $x^2+y^2=4$ drawn from the point $(4,0)$ are

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