Question

If the line $3x-4y-k=0,(k>0)$ touches the circle $x^2+y^2-4x-8y-5=0$ at $(a,b)$, then $k+a+b$ is equal to

• A. $20$
• B. $22$
• C. $-30$
• D. $-28$

Answer 1

The correct option is A $20$

Since the given line touches the given circle,
the length of the perpendicular from the center $(2,4)$ of the circle to the line

$3x-4y-k=0$ is equal to the radius $\sqrt{4+16+5}=5$ of the circle.
$\Rightarrow \frac{3\times 2-4\times 4-k}{\sqrt{9+16}}=\pm S\Rightarrow k=15$ $[\because k>0]$
Now, equation of the tangent at $(a,b)$ to the given circle is
$xa+yb-2(x+a)-4(y+b)-5=0$
$\Rightarrow (a-2)x+(b-4)y-(2a+4b+5)=0$
If it represents the given lines $3x-4y-k=0$
then, $\frac{a-2}{3}=\frac{b-4}{-4}=\frac{2a+4b+5}{k}=l$ (say)
Then, $a=3l+2,b=4-4l$ and $2a+4b+5=kl$ ...(1)
$\Rightarrow 2(3l+2)+4(4-4l)+5=15l$ $(\because k=15)$
$\Rightarrow l=1\Rightarrow a=5,b=0$ Then, $k+a+b=20$