<p>The equation of the circle which touches the axes of coordinates and the line $\frac{x}{3} + \frac{y}{4} = 1$ and whose centre lies in the first quadrant is $x^2 + y^2 - 2rx - 2ry + r^2 = 0$, then $r$ can be equal to:</p>
Step-by-Step Solution
Key Concept: A circle tangent to both axes in the first quadrant has centre $(r,r)$ and radius $r$; use the distance formula from centre to the given line.
<p><strong>Analysis:</strong> A circle touching both coordinate axes with centre in the first quadrant has centre at $(r, r)$ and radius $r$. The equation $x^2 + y^2 - 2rx - 2ry + r^2 = 0$ confirms this. For this circle to also touch the line $\frac{x}{3} + \frac{y}{4} = 1$ (or $4x + 3y = 12$), the distance from $(r,r)$ to this line must equal $r$: $$\frac{|4r + 3r - 12|}{\sqrt{16+9}} = r$$ $$\frac{|7r - 12|}{5} = r$$ $$|7r - 12| = 5r$$. This gives $7r - 12 = 5r$ (since $r > 0$ and we need $7r > 12$ in first quadrant) so $2r = 12$, thus $r = 6$.</p><p>∴ Answer is (d).</p>
Correct Answer: d