Question:
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
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
- \(12x-16y-15=0\)
- \(3x-4y+\dfrac{11}{2}=0\)
- \(12x-16y+15=0\)
- \(3x-4y-\dfrac{11}{2}=0\)
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The Correct Option is C
Solution and Explanation
Step 1: Understand the condition for the center of a circle.
If two straight lines are common tangents to a circle, then the center of the circle must be equidistant from both tangent lines.
Therefore, the locus of the center is the line midway between the two given parallel tangents.
Step 2: Write both lines in comparable form.
The given lines are \[ 3x-4y+4=0 \] and \[ 6x-8y+7=0 \] Dividing the second equation by \(2\), we get \[ 3x-4y+\frac{7}{2}=0 \]
Step 3: Find the middle line between two parallel lines.
The two parallel lines are \[ 3x-4y+4=0 \] and \[ 3x-4y+\frac{7}{2}=0 \] The middle line is obtained by taking the average of the constant terms.
So, the constant term of the locus is \[ \frac{4+\frac{7}{2}}{2} \] \[ =\frac{\frac{8}{2}+\frac{7}{2}}{2} \] \[ =\frac{\frac{15}{2}}{2} \] \[ =\frac{15}{4} \] Hence, the required locus is \[ 3x-4y+\frac{15}{4}=0 \]
Step 4: Convert into standard option form.
Multiplying by \(4\), we get \[ 12x-16y+15=0 \]
Step 5: Final conclusion.
Therefore, the locus of centers is \[ \boxed{12x-16y+15=0} \]
If two straight lines are common tangents to a circle, then the center of the circle must be equidistant from both tangent lines.
Therefore, the locus of the center is the line midway between the two given parallel tangents.
Step 2: Write both lines in comparable form.
The given lines are \[ 3x-4y+4=0 \] and \[ 6x-8y+7=0 \] Dividing the second equation by \(2\), we get \[ 3x-4y+\frac{7}{2}=0 \]
Step 3: Find the middle line between two parallel lines.
The two parallel lines are \[ 3x-4y+4=0 \] and \[ 3x-4y+\frac{7}{2}=0 \] The middle line is obtained by taking the average of the constant terms.
So, the constant term of the locus is \[ \frac{4+\frac{7}{2}}{2} \] \[ =\frac{\frac{8}{2}+\frac{7}{2}}{2} \] \[ =\frac{\frac{15}{2}}{2} \] \[ =\frac{15}{4} \] Hence, the required locus is \[ 3x-4y+\frac{15}{4}=0 \]
Step 4: Convert into standard option form.
Multiplying by \(4\), we get \[ 12x-16y+15=0 \]
Step 5: Final conclusion.
Therefore, the locus of centers is \[ \boxed{12x-16y+15=0} \]
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