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 $3 x-4 y+6=0$
$\Rightarrow h+k=5$
