Question

A tangent PT is drawn to the circle ${{x}^{2}}+{{y}^{2}}=4$ at the point $P\left( \sqrt{3},1 \right)$. A straight-line L, perpendicular to PT is a tangent to the circle ${{\left( x-3 \right)}^{2}}+{{y}^{2}}=1$. A possible equation of equation L is
A. $x-\sqrt{3}y=1$
B. $x+\sqrt{3}y=1$
C. $x-\sqrt{3}y=-1$
D. $x+\sqrt{3}y=5$

Answer 1

Hint: We first try to find the slope of the tangent PT and then using the slope of perpendicular lines we find the slope of the tangent to the circle ${{\left( x-3 \right)}^{2}}+{{y}^{2}}=1$. We use the simplification formula of $y=mx\pm \sqrt{{{m}^{2}}+{{a}^{2}}}$ to find the possible equation of the tangent.

Complete step by step solution:
A tangent PT is drawn to the circle ${{x}^{2}}+{{y}^{2}}=4$ at the point $P\left( \sqrt{3},1 \right)$. We know that equation of tangent at point $\left( h,k \right)$ for circle ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ is $xh+yk={{a}^{2}}$.
Therefore, the tangent PT will be $\sqrt{3}x+y=4$. The equation can be written as $y=-\sqrt{3}x+4$ which gives the slope as $m=-\sqrt{3}$ because the general equation of line is $y=mx+c$.
A straight-line L, perpendicular to PT is a tangent to the circle ${{\left( x-3 \right)}^{2}}+{{y}^{2}}=1$. Let the slope of the line L be ${{m}_{1}}$.
We know that the multiplication of slopes of two perpendicular line is always equal to $-1$.
Therefore, $m{{m}_{1}}=-1$. We have $m=-\sqrt{3}$ which gives ${{m}_{1}}=\dfrac{1}{\sqrt{3}}$.
Now for equation ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ and its tangent the simplification gives $y=mx\pm \sqrt{{{m}^{2}}+{{a}^{2}}}$ where m is the slope for the tangent $y=mx+c$.
Putting the values for ${{\left( x-3 \right)}^{2}}+{{y}^{2}}=1$ and the slope of its tangent ${{m}_{1}}=\dfrac{1}{\sqrt{3}}$, we can have
$y=m\left( x-3 \right)\pm \sqrt{{{m}^{2}}+{{a}^{2}}}=\dfrac{1}{\sqrt{3}}\left( x-3 \right)\pm \sqrt{\dfrac{1}{3}+1}=\dfrac{\left( x-3 \right)\pm 2}{\sqrt{3}}$.
Now we find the two possible slopes by simplifying which are $x-\sqrt{3}y=1$ or $x-\sqrt{3}y=5$.
The correct option is A.

Note: We have to be careful about the change of the origin of the circle. The simplification formula was for ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$. For our given circle the centre is at $\left( 3,0 \right)$. Therefore, the formula changed to $y=m\left( x-3 \right)\pm \sqrt{{{m}^{2}}+{{a}^{2}}}$.