Question:

The radius of the circle having \[ 3x-4y+4=0 \] and \[ 6x-8y-7=0 \] as its tangents is

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For a circle touching two parallel lines, the diameter of the circle equals the perpendicular distance between the lines.
Updated On: Jun 22, 2026
  • \(\frac32\)
  • \(3\)
  • \(6\)
  • \(\frac34\)
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The Correct Option is D

Solution and Explanation

Step 1: Observe the given lines.
The lines are \[ 3x-4y+4=0 \] and \[ 6x-8y-7=0 \] Divide the second equation by \(2\): \[ 3x-4y-\frac72=0 \] Thus, the two lines are parallel.

Step 2: Use the property of tangents to a circle.
If two parallel lines are tangents to a circle, then the distance between the lines equals the diameter of the circle.
Hence, \[ \text{Radius}=\frac12\times(\text{distance between the lines}) \]

Step 3: Find the distance between the parallel lines.
The lines are: \[ 3x-4y+4=0 \] and \[ 3x-4y-\frac72=0 \] Distance between parallel lines \[ ax+by+c_1=0 \] and \[ ax+by+c_2=0 \] is \[ \frac{|c_1-c_2|}{\sqrt{a^2+b^2}} \] Thus, \[ d= \frac{\left|4-\left(-\frac72\right)\right|}{\sqrt{3^2+(-4)^2}} \] \[ = \frac{\left|4+\frac72\right|}{\sqrt{9+16}} \] \[ = \frac{\frac{15}{2}}{5} \] \[ =\frac32 \]

Step 4: Find the radius.
Since diameter \[ =\frac32, \] radius is \[ r=\frac12\times\frac32 \] \[ =\frac34 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\frac34} \]
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