3x+4y-2=0 and 6x+8y-4=o are may have

1.	3x+4y-2=0 and 6x+8y-4=o are may have
Answer» Step-by-step explanation: The given lines being parallel tangents to a circle, the diameter of the circle is EQUAL to the distance between these lines, So that the required radius is 2 1 × 9+16 4+ 2 7 = 2 1 × 2 15 × 5 1 = 4 3 The center of the circle LIES on the LINE parallel to the given lines at a distance of 4 3 from each of them. So let the EQUATION be 3x−4y+k=0 ...(1) then 9+16 k−4 =± 4 3 ⇒k=4±( 4 15 )⇒k= 4 1 or 4 31 For k= 4 1 , distance of (1) from the other line is also 4 3 . Thus the center lies on the line 12x−16y+1=0

Answer»

Step-by-step explanation:

The given lines being parallel tangents to a circle,

the diameter of the circle is EQUAL to the distance between these lines,

So that the required radius is

9+16

The center of the circle LIES on the LINE parallel to the given lines at a distance of

from each of them.

So let the EQUATION be 3x−4y+k=0 ...(1)

then

9+16

k−4

=±

⇒k=4±(

)⇒k=

For k=

, distance of (1) from the other line is also

Thus the center lies on the line 12x−16y+1=0

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