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

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10-1.Circle and System of Circles

hard

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 :

A

$5$

B

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

C

$6$

D

$5 \sqrt{2}$

(JEE MAIN-2023)

Solution

$L _1: y=x+2, L_2: 4 y=3 x+6, L_3: 3 y=4 x+1$

Bisector of lines $L _2 \& L _3$

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

$\text { (+) } 4 x-3 y+1=3 x-4 y+6$

$x+y=5$

Centre lies on Bisector of $4 x-3 y+1=0 \&$

$\text { (0) } 3 x-4 y+6=0$

$\Rightarrow h + k =5$

Standard 11

Mathematics

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