A tangent PT is drawn to the circle x^2+y^2=4 | vedclass

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(A) $1.$ The equation of the tangent to $x^2+y^2=4$ at $P(\sqrt{3}, 1)$ is $\sqrt{3}x + y = 4$.The centers of the circles are $C_1(0,0)$ with radius $

Vedclass · Accepted Answer

$(D, A)$

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(A) $1.$ The equation of the tangent to $x^2+y^2=4$ at $P(\sqrt{3}, 1)$ is $\sqrt{3}x + y = 4$.The centers of the circles are $C_1(0,0)$ with radius $r_1=2$ and $C_2(3,0)$ with radius $r_2=1$.The external center of similitude $B$ divides $C_1C_2$ in the ratio $r_1:r_2 = 2:1$ externally.$B = \left(\frac{2(3)-1(0)}{2-1}, \frac{2(0)-1(0)}{2-1}\right) = (6,0)$.Let the common tangent be $y-0 = m(x-6)$,or $mx - y - 6m = 0$.The perpendicular distance from $C_1(0,0)$ to this line equals $r_1=2$.$\left|\frac{-6m}{\sqrt{m^2+1}}\right| = 2$ $\Rightarrow 3|m| = \sqrt{m^2+1}$ $\Rightarrow 9m^2 = m^2+1$ $\Rightarrow m^2 = \frac{1}{8}$ $\Rightarrow m = \pm \frac{1}{2\sqrt{2}}$.The equations are $x \pm 2\sqrt{2}y = 6$. Thus,$(D)$ is a common tangent.$2.$ The slope of $PT$ $(\sqrt{3}x+y=4)$ is $m_1 = -\sqrt{3}$.Since $L$ is perpendicular to $PT$,the slope of $L$ is $m_2 = -\frac{1}{m_1} = \frac{1}{\sqrt{3}}$.Let the equation of $L$ be $x - \sqrt{3}y + k = 0$.$L$ is tangent to $(x-3)^2+y^2=1$,so the distance from $(3,0)$ to $L$ is $1$.$\left|\frac{3 - \sqrt{3}(0) + k}{\sqrt{1^2 + (-\sqrt{3})^2}}\right| = 1 \Rightarrow \left|\frac{3+k}{2}\right| = 1$.$3+k = 2$ or $3+k = -2$,so $k = -1$ or $k = -5$.The possible equations are $x - \sqrt{3}y - 1 = 0$ or $x - \sqrt{3}y - 5 = 0$.Comparing with options,$x - \sqrt{3}y = 1$ is not listed,but $x - \sqrt{3}y = -1$ is option $(C)$.

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