Circle Question 6

Question 6 - 30 January - Shift 1

Let $y=x+2,4 y=3 x+6$ and $3 y=4 x+1$ be three tangent lines to the circle $(x-h)^{2}+(y-k)^{2}=r^{2}$. Then $h+k$ is equal to :

(1) 5

(2) $5(1+\sqrt{2})$

(3) 6

(4) $5 \sqrt{2}$

Show Answer

Answer: (1)

Solution:

Formula: Bisectors Of The Angles Between Two Lines:

$L_1: y=x+2, $

$L_2: 4 y=3 x+6, $

$ L_3: 3 y=4 x+1$

Bisector of lines $L_2 $ and $ L_3$

$ \frac{4 x-3 y+1}{5}= \pm(\frac{3 x-4 y+6}{5}) $

Taking positive sign, we get

$\Rightarrow 4 x-3 y+1=3 x-4 y+6$

$ \Rightarrow x+y=5 $

Centre lies on Bisector of $4 x-3 y+1=0 $ and $3 x-4 y+6=0$

$\Rightarrow h+k=5$