The Locus of centers of the circles, possessing the same area and having 3 x - 4 y + 4 = 0 and 6 x - 8 y - 7…

Given,

$3x-4y+4=0$ and $6x-8y-7=0$ are common tangent,

Now the given lines are parallel tangents to a circle,

So, the diameter of the circle is equal to the distance between these lines,

So that the required radius is

$\frac{1}{2}\times \frac{4+\frac{7}{2}}{\sqrt{9+16}}=\frac{1}{2}\times \frac{15}{2}\times \frac{1}{5}=\frac{3}{4}$

The center of the circle lies on the line parallel to the given lines at a distance of $\frac{3}{4}$ from each of them.

So let the equation passing from center be $3x-4y+k=0 \cdots \left(1\right)$

Then by distance between two parallel line formula we get, $\frac{k-4}{\sqrt{9+16}}=\pm \frac{3}{4}\Rightarrow k=4\pm \left(\frac{15}{4}\right)\Rightarrow k=\frac{1}{4}$ or $\frac{31}{4}$

For $k=\frac{1}{4}$, distance of $\left(1\right)$ from the other line is also $\frac{3}{4}$.

Thus the center lies on the line $3x-4y+\frac{1}{4}=0$

 $\Rightarrow 12x-16y+1=0$