Question

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. $5 \sqrt{2}$

Answer 1

Answer: (A)

5

Answer 2

L1​:y=x+2,L2​:4y=3x+6,L3​:3y=4x+1
Bisector of lines L2​&L3​
54x−3y+1​=±(53x−4y+6​)
(+)4x−3y+1=3x−4y+6
x+y=5
Centre lies on Bisector of 4x−3y+1=0&
(0) 3x−4y+6=0
⇒h+k=5