If two tangents are drawn from the point P on | vedclass

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(C) The equation of the circle is $x^2+y^2=1$. The radius $r=1$ and the center is $(0,0)$.Let the slope of the tangent be $m$. The equation of a line

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$2 \pm \sqrt{3}$

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(C) The equation of the circle is $x^2+y^2=1$. The radius $r=1$ and the center is $(0,0)$.Let the slope of the tangent be $m$. The equation of a line passing through $P(\sqrt{2}, \sqrt{2})$ with slope $m$ is $y - \sqrt{2} = m(x - \sqrt{2})$,which simplifies to $mx - y + \sqrt{2}(1-m) = 0$.Since this line is a tangent to the circle $x^2+y^2=1$,the perpendicular distance from the center $(0,0)$ to the line must be equal to the radius $r=1$.Using the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2+b^2}}$,we get $1 = \frac{|m(0) - 1(0) + \sqrt{2}(1-m)|}{\sqrt{m^2+(-1)^2}}$.This implies $1 = \frac{|\sqrt{2}(1-m)|}{\sqrt{m^2+1}}$.Squaring both sides,we get $m^2+1 = 2(1-m)^2$.$m^2+1 = 2(1 - 2m + m^2) = 2 - 4m + 2m^2$.Rearranging the terms,we get $m^2 - 4m + 1 = 0$.Using the quadratic formula $m = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $m = \frac{4 \pm \sqrt{16-4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$.

If two tangents are drawn from the point $P$ on the circle $x^2+y^2=4$ to the circle $x^2+y^2=1$,where the point $P$ is given by $(\sqrt{2}, \sqrt{2})$,then the slopes of the tangents are:

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