Question

if the equation of tangent to the circle $x^2+y^2-6x+4y-12=0$ which are parallel to the line 4x+3y+5=0 is ax+by+c=0,find the value of $-(\frac{a}{b}+\frac{c}{b})$, where $\left|\frac{c}{a}\right|>6$

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Answer 1

We have to find the tangent parallel to the line 4x+3y+5=0. We know that any line parallel to 4x+3y+5=0 can be written as 4x+3y+k=0.Let it be the equation of our tangent.

similar to previous problems, we will use the condition that the perpandicular distance form centre to the tangent is equal to the radius to find k.

$\Rightarrow distance$ of (3,-2) from 4x+3y+k= radius

$\Rightarrow \left|\frac{4\times 3+3x-2+k}{\sqrt{4^2+3^2}}\right|=\sqrt{3^2+2^2-(-12)}$

$\Rightarrow \left|\frac{k+6}{5}\right|=5$

$\Rightarrow |k+6|=25$

$\Rightarrow k+6=25$

$\Rightarrow k+6=25$

$\Rightarrow k=19or-31$

So the equation of tangent are 4x+3y+19 and 4x+3y-31=0

We are given one more condition that $\left|\frac{c}{a}\right|>6$. only 4x+3y-31 satisfies this condition.

$\Rightarrow$ 4x+3y-31=0 is the realized equation 0r 4x+3y-31=0 $\equiv$ax+by+c=0

We want find $-(\frac{a}{b}+\frac{c}{b})=-(\frac{4}{3}+\frac{-31}{3})$

=9