Question

Find the radius of the circle that has the lines: $$ 3x - 4y + 4 = 0 \quad \text{and} \quad 6x - 8y - 7 = 0 $$ as its tangents.

• A. \( \frac{3}{2} \)
• B. \( 3 \)
• C. \( 6 \)
• D. \( \frac{3}{4} \)

Hint:

For circles with two parallel tangents, use the perpendicular distance formula and divide by 2 to get the radius.

Answer 1

The Correct Option is D

The circle has two tangents, so the radius is half the distance between parallel tangents. Check whether the given lines are parallel: Lines: - \( L_1: 3x - 4y + 4 = 0 \) - \( L_2: 6x - 8y - 7 = 0 \) Simplify \( L_2 \) by dividing by 2: \[ 3x - 4y - \frac{7}{2} = 0 \] Now both lines are in the form \( 3x - 4y + c = 0 \), so they are parallel. Now, distance between parallel lines: \[ \text{Distance} = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = \frac{|4 - (-\frac{7}{2})|}{\sqrt{3^2 + (-4)^2}} = \frac{4 + \frac{7}{2}}{5} = \frac{\frac{15}{2}}{5} = \frac{3}{2} \] Radius = half of this: \[ \Rightarrow R = \frac{1}{2} \cdot \frac{3}{2} = \boxed{\frac{3}{4}} \]