Common tangents which are parallel are ‌3x−4y+4=0 and 6x−8y−7=0 ⇒3x−4y+4=0 and 3x−4y−‌
7
2
=0 The diameter is equal to the distance between these parallel lines. So, the radius ‌=‌
1
2
×|‌
4−(−
7
2
)
√32+(−4)2
| ‌=‌
1
2
×‌
15
2
×‌
1
5
=‌
3
4
The centre of the circle lies on the line parallel to the given lines at a distance of ‌
3
4
from both the lines. So, let the equation be 3x−4y+k=0.....(i) Then, distance between 3x−4y+k=0 and 3x−4y+4 is same as distance between 3x−4y+k=0‌ and ‌3x−4y−‌
7
2
=0
‌|‌
k−4
√32+(−4)2
|=|‌
k−(−
7
2
)
√(3)2+(−4)2
|
‌⇒|k−4|=−|k+‌
7
2
| and −(k−4)=k+‌
7
2
∴‌‌2k=‌
7
2
−4=−‌
1
2
⇒k=−‌
1
4
and 2k=4−‌
7
2
=‌
1
2
⇒k=‌
1
4
∴ For k=‌
1
4
distance of Eq. (i) from other line is ‌
3
4
∴ Locus of centers of the circles, is 3x−4y+‌