1. Write the distance from origin

For a point on the curve $x=a\sin t-b\sin\left(\frac{at}{b}\right), \qquad y=a\cos t-b\cos\left(\frac{at}{b}\right),$ the distance from origin is $r=\sqrt{x^2+y^2}.$

So, $r^2=x^2+y^2.$

2. Compute $x^2+y^2$

Let $\alpha=t, \qquad \beta=\frac{at}{b}.$ Then $x=a\sin\alpha-b\sin\beta, \qquad y=a\cos\alpha-b\cos\beta.$

Hence, \begin{align*} r^2 &= (a\sin\alpha-b\sin\beta)^2+(a\cos\alpha-b\cos\beta)^2 \ &= a^2(\sin^2\alpha+\cos^2\alpha)+b^2(\sin^2\beta+\cos^2\beta) \ &\quad -2ab(\sin\alpha\sin\beta+\cos\alpha\cos\beta). \end{align*}

Using $\sin^2\theta+\cos^2\theta=1$ and $\sin\alpha\sin\beta+\cos\alpha\cos\beta=\cos(\alpha-\beta),$ we get $r^2=a^2+b^2-2ab\cos(\alpha-\beta).$

Now, $\alpha-\beta=t-\frac{at}{b}=t\left(1-\frac{a}{b}\right).$ So, $r^2=a^2+b^2-2ab\cos\left[t\left(1-\frac{a}{b}\right)\right].$

3. Find the maximum value

Since $-1\le \cos\left[t\left(1-\frac{a}{b}\right)\right]\le 1,$ we have $r^2=a^2+b^2-2ab\cos(\cdots).$

This is maximum when $\cos(\cdots)=-1.$ Therefore, $r^2_{\max}=a^2+b^2+2ab=(a+b)^2.$

Thus, $r_{\max}=a+b,$ since $a,b>0$.

4. Check options

• A: $a-b$ → not maximum
• B: $a+b$ → correct
• C: $\sqrt{a^2+b^2}$ → not maximum
• D: $\sqrt{a^2-b^2}$ → not correct in general

Therefore, the correct option is B.