Question

5. The polar of (2,-1) w.r.t x^2+y^2+ 6x + 4y-1=0 is 5x + y + k=0 then k=​

Answer 1

Step-by-step explanation:

5x+y+k=0 (put x=2; y=-1)

5×2+(-1)+k=0

10-1+k=0

9+k=0

k=-9

Answer 2

Given: The polar of $(2,-1)$ w.r.t $x^2+y^2+ 6x + 4y-1=0$ is $5x + y + k=0$

To Find: we have to find the value of k.

Step by step Solution:

Here we have $(2,-1)$ we get x and y $x=2, y=-1$

Put in the equation containing k

$5x+y+k=0\\5×2+(-1)+k=0\\10-1+k=0\\9+k=0\\k=-9$

Final answer:

Hence the value of k is -9