Given one tangent $4x-y=0$ from passing through origin and to the circle with centre (2,4) . What is the equation of the other tangent through origin
I have tried by finding the radius but the process is taking too much time pls recommend a shorter one
As the radius $r=$ the perpendicular distance from the center,
$r =\dfrac{|4\cdot2-1\cdot4|}{\sqrt{4^2+(-1)^2}}$
If $y=mx$ is another tangent, $r=\dfrac{|2m-4|}{\sqrt{1+m^2}}$
Your radius will be on a line perpendicular to the tangent line, passing through the centre. You can get the slope of the radius by taking the negative reciprocal of the tangent's slope. You have a point. There's your line. Where does it intersect with the tangent? That's a point on the circle. You now have a point on the circle and the centre point. Use the distance formula, and you have your radius. Radius. Centre. Plug it into your circle equation. The other tangent line will have the same length as the first, between origin and point of contact. That should help you fix your 2nd tangent.
Let $k=2$ be the slope of the line from the origin to the center; let $k_1= 4$ and $k_2$ be the slopes of the two tangent lines. Since the circle center line bisects the angle between the two tangent lines,
$$ \frac{k_1-k}{1+k_1k}= \frac{k-k_2}{1+k_2k} \implies \frac{4-2}{1+4\cdot 2}= \frac{2-k_2}{1+2\cdot k_2}$$
Thus, the other tangent line is $y=k_2x=\frac{16}{13}x$.
The given tangent line (shown in blue) through the origin has slope $4$, so it makes an angle $\arctan4$ with the $x$-axis. The line (shown in violet) through the origin and the center of the circle (shown in green) has slope $2$, so it makes an angle $\arctan2$ with the $x$-axis. The difference between the angle $\theta$ of the other tangent line (shown in red) through the origin to angle of the line of slope $2$ equals the angle from the line of slope $2$ to the given tangent line with slope $4$. That is, $\arctan4-\arctan2=\arctan2-\theta$, so $\theta=2\arctan2-\arctan4$, so $\tan\theta=\frac{16}{13}$, so the other tangent line is $y=\frac{16}{13}x$.
It shouldn’t take very long to compute the radius since it’s simply the distance between the given center and tangent line, for which there’s a standard formula. However, you don’t need to know the circle’s radius in order to solve this problem. The tangents through an exterior point to a circle are symmetric with respect to the line through that point and the circle’s center, so the other tangent is the reflection of $4x-y=0$ in the line $y=2x$.
A normal vector to the line being reflected is $(4,-1)$ and a normal to the reflector is $(2,-1)$, so using a well-known formula for reflection of a vector, a normal to the tangent line’s reflection is $$(4,-1)-2{(4,-1)\cdot(2,-1)\over(2,-1)\cdot(2,-1)}(2,-1) = \left(-\frac{16}5,\frac{13}5\right),$$ therefore an equation of the other tangent line is $16x=13y$.
