If $$\theta$$ is the angle between the pair of tangents drawn from the point $$\left(0, \frac{7}{2}\right)$$ to the circle $$x^2 + y^2 - 14x + 16y + 88 = 0$$, then $$\tan \theta$$ equals

The equation of the circle reduces to $$(x-7)^2+(y+8)^2=5^2$$ when we add $$49+64$$ to both sides and subtract $$88$$ from both sides to complete the squares. This means that the centre of the circle is at $$(7,-8)$$ and its radius is $$5$$ units.

Using the Distance Formula, we get the distance between the centre and the point where tangents originate (say $$P$$) as:

$$\sqrt{(7-0)^2 + (-8-3.5)^2} = \dfrac{5\sqrt{29}}{2}$$

Along with the radius of the circle in question, the tangent and this distance between the centre and the point $$P$$ form a right angled triangle. Therefore, the length of each of the tangents will be $$\sqrt{\left(\dfrac{5\sqrt{29}}{2}\right)^2 - 5^2} = \dfrac{25}{2}$$.

$$\text{tan }\dfrac{\theta}{2}$$ is therefore equal to $$\dfrac{5}{\frac{25}{2}} = \dfrac{2}{5}$$

Using the formula $$\text{tan }2x = \dfrac{2\text{ tan }x}{1-\text{tan }^2 x}$$, we obtain $$\text{tan }\theta$$ as:

$$\dfrac{2*\frac{2}{5}}{1-\frac{4}{25}} = \dfrac{20}{21}$$