$\text{ The line }{L}_1\equiv 4x+3y-12=0\text{ intersects the }x-\text{ and }y\text{ -axis at }A\text{ and }B,\text{ respectively. A varıable ine }$

$\text{ perpendicular to }{L}_1\text{ intersects the }x\text{ - and the }y\text{ -axis at }P\text{ and }Q\text{ , respectively. Then the locus of the circumcenter of triangle }ABQ\text{ is }$

Clearly, the circumcenter of triangle ABQ will lie on the perpendicular bisector of line AB.

If A(3,0) and B(0,4) then midpoint of AB =$\left(\frac{3}{2},2\right)$ 

Let the equation of perpendicular bisector be 3x-4y+k=0, which passes through $\left(\frac{3}{2},2\right)$

Now, the equation of perpendicular bisector of line $AB\text{ is }3x-4y+\frac{7}{2}=0\text{ . Hence, }$

$\text{ the locus of circumcenter is }6x-8y+7=0$