Let $m_1$ and $m_2$ are the slopes of the tangents drawn to circle $x^2+y^2-4x-8y-5=0$ from the point $P(-1,-2)$,and $|m_1+m_2|=\frac{p}{q}$

Question

Let $m_1$ and $m_2$ are the slopes of the tangents drawn to circle $x^2+y^2-4x-8y-5=0$ from the point $P(-1,-2)$,and $|m_1+m_2|=\frac{p}{q}$,where $p$and $q$ are relatively prime natural numbers,then find $p+q.$

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Slope of $OP=m=\frac{2}{1}$

$\tan\theta=\frac{|m_1-2|}{|1+2m_1|}$ and $\tan\theta=\frac{|m_2-2|}{|1+2m_2|}$
$\frac{|m_1-2|}{|1+2m_1|}=\frac{|m_2-2|}{|1+2m_2|}$
$|m_1-2||1+2m_2|=|m_2-2||1+2m_1|$
$|(m_1-2)(1+2m_2)|=|(m_2-2)(1+2m_1)|$
Taking plus and minus signs,we get

either $m_1=m_2$ or $4-4m_1m_2+3m_1+3m_2=0$

But i am not able to find the value of $|m_1+m_2|$.Please help me.

Answer 1

Note that $OP=3\sqrt 5$ and that the radius is $5$. So, you can have $$\tan\theta=\frac{5}{\sqrt{(3\sqrt 5)^2-5^2}}=\frac{\sqrt 5}{2}.$$

So, you have $$\frac{|m-2|}{|1+2m|}=\frac{\sqrt 5}{2}.$$ Squaring the both sides gives $$16m^2+36m-11=0.$$ So, by Vieta's formulas, $$|m_1+m_2|=\left|-\frac{36}{16}\right|=\frac 94.$$