If the line 3x -4y -k = 0 (k 0) touches | vedclass

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Question

since, the given line touches the given circle, the length of the perpendicular from the centre $(2,4)$ of the circle to the line $3 x-4 y-k=0$ is equ

Vedclass · Accepted Answer

$20$

Vedclass · Answer

since, the given line touches the given circle, the length of the perpendicular from the centre $(2,4)$ of the circle to the line $3 x-4 y-k=0$ is equal

to the radius $\sqrt{4+16+5}=5$ of the circle.

$	herefore \frac{3 	imes 2-4 	imes 4-k}{\sqrt{9+16}}=\pm 5$

$\Rightarrow \quad k=15$              $[\because k>0]$

hence equation of tangent is

$3 x-4 y-15=0$          .......$(1)$

Let equation of normal to circle

$4 x+3 y=\lambda$

It passes through centre $(2,4)$

$\Rightarrow \lambda=20$

hence equation of normal is

$4 x+3 y=20$             .........$(2)$

Solve $( 1)$ and $(2)$

$a=5, \quad b=0$

$k+a+b=15+5+0=20$

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 :-

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