Given that we need to find the equation of the circle with centre (3, 4) and touches the straight line 5x + 12y - 1 = 0.

Since the circle touches the line at a single point and the circle passes through that point.

We know that the radius of a circle is the distance between the centre and any point that lies on the circle.

Here the point lies on the circle and also on the line, the distance between the points is equal to the perpendicular distance from the centre on to the line 5x + 12y - 1 = 0.

We know that the perpendicular distance from the point (x₁,y₁) on to the line ax + by + c = 0 is given by .

Let us assume ‘r’ be the radius of the circle.

⇒

⇒

⇒

⇒ .

We have a circle with centre (3, 4) and having a radius .

We know that the equation of the circle with centre (p, q) and having radius ‘r’ is given by:

⇒ (x - p)² + (y - q)² = r²

Now we substitute the corresponding values in the equation:

⇒

⇒

⇒ 169x² + 169y² - 1014x - 1352y + 4225 = 3844

⇒ 169x² + 169y² - 1014x - 1352y + 381 = 0.

∴The equation of the circle is 169x² + 169y² - 1014x - 1352y + 381 = 0.