1. State the problem: Solve for $r$ in the equation $$r^2 + (r + 5)^2 = 17^2.$$\n\n2. Expand the squared binomial: $$(r + 5)^2 = r^2 + 2 \cdot r \cdot 5 + 5^2 = r^2 + 10r + 25.$$\n\n3. Substitute back into the equation: $$r^2 + r^2 + 10r + 25 = 289,$$ since $17^2 = 289.$\n\n4. Combine like terms: $$2r^2 + 10r + 25 = 289.$$\n\n5. Subtract 289 from both sides: $$2r^2 + 10r + 25 - 289 = 0,$$ which simplifies to $$2r^2 + 10r - 264 = 0.$$\n\n6. Divide the entire equation by 2 to simplify: $$r^2 + 5r - 132 = 0.$$\n\n7. Factor the quadratic: We look for two numbers that multiply to $-132$ and add to $5$. Those numbers are $12$ and $-11$. So, \n$$r^2 + 5r - 132 = (r + 12)(r - 11) = 0.$$\n\n8. Set each factor equal to zero and solve for $r$: \n$$r + 12 = 0 \implies r = -12,$$ and $$r - 11 = 0 \implies r = 11.$$\n\n9. Since $r$ represents a radius (a length) in the geometric context, it must be non-negative. Therefore, we discard $r = -12$ and keep $$r = 11.$$\n\nFinal answer: $$\boxed{11}.$$