An angle of intersection of the curves, $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ and $$x^2 + y^2 = ab$$, $$a \gt b$$, is:

We have to find the acute angle at which the ellipse $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$ and the circle $$x^{2}+y^{2}=ab$$ intersect, the given condition being $$a\gt b.$$

First we locate the points of intersection. Any common point must satisfy both equations, so from the second curve we write

$$x^{2}+y^{2}=ab\;.$$

We substitute $$x^{2}=ab-y^{2}$$ in the ellipse equation:

$$\frac{ab-y^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1.$$

Multiplying every term by $$a^{2}b^{2}$$ to clear the denominators,

$$b^{2}(ab-y^{2})+a^{2}y^{2}=a^{2}b^{2}.$$

Expanding the left side,

$$ab\,b^{2}-b^{2}y^{2}+a^{2}y^{2}=a^{2}b^{2}.$$

Collecting all terms on one side,

$$ab\,b^{2}-a^{2}b^{2}+y^{2}(a^{2}-b^{2})=0.$$

Factorising the constant part,

$$-a b^{2}(a-b)+y^{2}(a^{2}-b^{2})=0.$$

The factor $$a^{2}-b^{2}=(a-b)(a+b)$$ is positive because $$a\gt b,$$ so we solve for $$y^{2}:$$

$$y^{2}=\frac{a b^{2}(a-b)}{(a-b)(a+b)}=\frac{a b^{2}}{a+b}.$$

Using $$x^{2}=ab-y^{2}$$ we get

$$x^{2}=ab-\frac{a b^{2}}{a+b}=ab\left(1-\frac{b}{a+b}\right)=ab\left(\frac{a}{a+b}\right)=\frac{a^{2}b}{a+b}.$$

Thus for every intersection point

$$x^{2}=\frac{a^{2}b}{a+b},\qquad y^{2}=\frac{a b^{2}}{a+b}.$$

Now we compute the slopes of the tangents to both curves at a common point.

Ellipse. Starting from $$F(x,y)=\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}-1=0,$$ differentiate implicitly:

$$\frac{2x}{a^{2}}+\frac{2y}{b^{2}}\frac{dy}{dx}=0.$$

Hence

$$\frac{dy}{dx}=-\frac{b^{2}x}{a^{2}y},$$

so the slope of the tangent to the ellipse is

$$m_{1}=-\frac{b^{2}x}{a^{2}y}.$$

Circle. From $$G(x,y)=x^{2}+y^{2}-ab=0$$ we differentiate:

$$2x+2y\frac{dy}{dx}=0\;\Longrightarrow\;\frac{dy}{dx}=-\frac{x}{y},$$

giving the circle’s tangent slope

$$m_{2}=-\frac{x}{y}.$$

The angle $$\theta$$ between two curves is the angle between their tangents, for which the standard relation is

$$\tan\theta=\left|\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right|.$$

We now evaluate each part at the intersection point.

First compute $$\dfrac{x}{y}.$$ From the already-found squares,

$$\left(\frac{x}{y}\right)^{2}=\frac{x^{2}}{y^{2}}=\frac{\dfrac{a^{2}b}{\,a+b\,}}{\dfrac{a b^{2}}{\,a+b\,}}=\frac{a^{2}b}{a b^{2}}=\frac{a}{b},$$

so

$$\Bigl|\frac{x}{y}\Bigr|=\sqrt{\frac{a}{b}}.$$

Next find $$m_{1}m_{2}:$$

$$m_{1}m_{2}=\left(-\frac{b^{2}x}{a^{2}y}\right)\left(-\frac{x}{y}\right)=\frac{b^{2}x^{2}}{a^{2}y^{2}}=\frac{b^{2}}{a^{2}}\cdot\frac{x^{2}}{y^{2}}=\frac{b^{2}}{a^{2}}\cdot\frac{a}{b}=\frac{b}{a}.$$

Therefore

$$1+m_{1}m_{2}=1+\frac{b}{a}=\frac{a+b}{a}.$$

Now evaluate $$m_{2}-m_{1}:$$

$$$ \begin{aligned} m_{2}-m_{1}&=-\frac{x}{y}-\left(-\frac{b^{2}x}{a^{2}y}\right) =-\frac{x}{y}+\frac{b^{2}x}{a^{2}y} =\frac{x}{y}\left(-1+\frac{b^{2}}{a^{2}}\right)\\ &=\frac{x}{y}\left(\frac{b^{2}-a^{2}}{a^{2}}\right) =-\frac{x}{y}\cdot\frac{a^{2}-b^{2}}{a^{2}}. \end{aligned} $$$

Taking absolute value removes the minus sign, so

$$|m_{2}-m_{1}|=\Bigl|\frac{x}{y}\Bigr|\frac{a^{2}-b^{2}}{a^{2}} =\sqrt{\frac{a}{b}}\;\frac{a^{2}-b^{2}}{a^{2}}.$$

Substituting everything into the formula for $$\tan\theta$$ gives

$$$ \begin{aligned} \tan\theta &=\frac{|m_{2}-m_{1}|}{1+m_{1}m_{2}} =\frac{\sqrt{\dfrac{a}{b}}\,\dfrac{a^{2}-b^{2}}{a^{2}}}{\dfrac{a+b}{a}} =\sqrt{\frac{a}{b}}\;\frac{a^{2}-b^{2}}{a^{2}}\;\frac{a}{a+b}. \end{aligned} $$$

Simplify the factors step by step. Notice $$a^{2}-b^{2}=(a-b)(a+b),$$ so

$$$ \begin{aligned} \tan\theta &=\sqrt{\frac{a}{b}}\;\frac{(a-b)(a+b)}{a^{2}}\;\frac{a}{a+b} =\sqrt{\frac{a}{b}}\;\frac{a-b}{a}. \end{aligned} $$$

Because $$\sqrt{\dfrac{a}{b}}\cdot\dfrac{1}{a} =\frac{\sqrt{a}}{\sqrt{b}}\cdot\frac{1}{a} =\frac{1}{\sqrt{ab}},$$ we finally reach

$$\;\tan\theta=\dfrac{a-b}{\sqrt{ab}}\;.$$

Therefore the required angle of intersection is $$\theta=\tan^{-1}\dfrac{a-b}{\sqrt{ab}}$$.

Hence, the correct answer is Option C.