Question

The locus of centers of the circles, possessing the same area and having $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ as their common tangent, is

• A. $12x - 16y - 15 = 0$
• B. $3x - 4y + \dfrac{11}{2} = 0$
• C. $12x - 16y + 15 = 0$
• D. $3x - 4y - \dfrac{11}{2} = 0$

Hint:

When circles are tangent to two parallel lines, their centers lie on the line equidistant from both.

Answer 1

The Correct Option is C

The given lines are parallel since their slopes are equal.
Any circle touching both must have its center equidistant from both lines.
Let the center be $(x, y)$. Distance from center to each line must be equal.
Use distance formula from point to line for both lines and equate:
$\left|\dfrac{3x - 4y + 4}{\sqrt{3^2 + 4^2}}\right| = \left|\dfrac{6x - 8y - 7}{\sqrt{6^2 + 8^2}}\right|$
Since both denominators simplify to same ratio (as lines are scalar multiples), equate numerators: $|3x - 4y + 4| = |6x - 8y - 7|$
Square both sides and simplify to find the required locus: $12x - 16y + 15 = 0$