Q.If the lines $3x - 4y + 4 = 0$ and $6x - 8y - 7 = 0$ are tangents to a circle, then find the radius of the circle.

Concept: Distance from a point to a line — for parallel tangents, the distance between them equals the diameter.

Step 1: Check parallelism.

The second line $6x - 8y - 7 = 0$ simplifies to $3x - 4y - 3.5 = 0$ (dividing by 2). Both lines have the same normal vector $(3, -4)$, so they are parallel.

Step 2: Distance between two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is

Step 3: Here $c_1 = 4$, $c_2 = -3.5$, $a = 3$, $b = -4$. …

Why this approach works

When two lines are both tangents to the same circle, they must be parallel — because a circle can only have two parallel tangents (one on each side). The distance between these two parallel tangents is exactly the diameter of the circle. So the problem reduces to: find the distance between two parallel lines, then halve it.

Let’s check if the lines are indeed parallel.

Step-by-step solution

1. Check if the lines are parallel

Line 1: $3x - 4y + 4 = 0$

Line 2: $6x - 8y - 7 = 0$

Divide the second equation by 2:

$3x - 4y - \frac{7}{2} = 0$

Now both have the same coefficients for $x$ and $y$ ($3$ and $-4$), so they are parallel. The only difference is the constant term.

2. Write both lines in the standard parallel form

Line 1: $3x - 4y + 4 = 0$

Line 2: $3x - 4y - \frac{7}{2} = 0$

Here $a = 3$, $b = -4$, $c_1 = 4$, $c_2 = -\frac{7}{2}$.

3. Apply the distance formula for parallel lines

Distance $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$

First, $c_1 - c_2 = 4 - \left(-\frac{7}{2}\right) = 4 + \frac{7}{2} = \frac{8}{2} + \frac{7}{2} = \frac{15}{2}$

So $|c_1 - c_2| = \frac{15}{2}$

Now $\sqrt{a^2 + b^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ …