Question

3x+4y-2=0 and 6x+8y-4=o are may have​

Answer 1

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