Question:medium

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

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The center of a circle tangent to two parallel lines lies on the line exactly midway between those two parallel lines.
Updated On: Jun 15, 2026
  • \(12x-16y-15=0\)
  • \(3x-4y+\dfrac{11}{2}=0\)
  • \(12x-16y+15=0\)
  • \(3x-4y-\dfrac{11}{2}=0\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understand the geometric condition.
If two parallel lines are both tangent to a circle, the centre is equidistant from each, so it lies on the line exactly midway between them.
Step 2: Write both tangents with the same coefficients.
The first is $3x-4y+4=0$. Dividing $6x-8y+7=0$ by $2$ gives $3x-4y+\dfrac72=0$. They are parallel.
Step 3: Take the midway line.
The locus is $3x-4y+k=0$ where $k$ is the average of the two constants.
Step 4: Average the constants.
$k=\dfrac{4+\frac72}{2}=\dfrac{\frac{15}{2}}{2}=\dfrac{15}{4}$.
Step 5: Write the locus.
So the locus is $3x-4y+\dfrac{15}{4}=0$.
Step 6: Clear the fraction.
Multiplying by $4$: $12x-16y+15=0$.
\[ \boxed{12x-16y+15=0} \]
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