Match each entry in List-I to the correct entry in List-II and choose the correct option.

Evaluating entry (P):

Radius $$R$$ equals the perpendicular distance from center $$(1,2)$$ to $$3x + 4y - 1 = 0$$:

$$R = \frac{\vert{}3(1) + 4(2) - 1\vert{}}{\sqrt{3^2 + 4^2}} = \frac{10}{5} = 2$$

$$\text{Equation of circle: } (x-1)^2 + (y-2)^2 = 4$$

Testing point (3,2) from List-II: $$(3-1)^2 + (2-2)^2 = 4 \implies 4 = 4 \implies \text{(P)} \rightarrow \text{(3)}$$

Evaluating entry (Q):

Equation of tangent to $$y^2 = 8x$$ with slope $$m$$: $$y = mx + \frac{2}{m}$$

Since it touches $$x^2 + y^2 = 2$$: $$\frac{\vert{}2/m\vert{}}{\sqrt{1 + m^2}} = \sqrt{2} \implies \frac{4}{m^2(1+m^2)} = 2 \implies m^4 + m^2 - 2 = 0 \implies m = 1 \quad (m > 0)$$

$$\text{Tangent line: } y = x + 2$$

Testing point (7,9) from List-II: $$9 = 7 + 2 \implies \text{(Q)} \rightarrow \text{(2)}$$

Evaluating entry (R):

$$\frac{x^2}{16} + \frac{y^2}{12} = 1 \implies a=4, \ b=\sqrt{12}$$

$$e = \sqrt{1 - \frac{12}{16}} = \frac{1}{2} \implies ae = 2$$

$$M = \left(ae, \frac{b^2}{a}\right) = \left(2, \frac{12}{4}\right) = (2, 3)$$

Equation of normal at $$M(2,3)$$: $$\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2 \implies \frac{16x}{2} - \frac{12y}{3} = 16 - 12 \implies 2x - y = 1$$

Testing point (1,1) from List-II: $$2(1) - 1 = 1 \implies \text{(R)} \rightarrow \text{(1)}$$

Evaluating entry (S):

$$ae = 5, \quad \frac{a}{e} = \frac{16}{5} \implies a^2 = 16 \implies b^2 = a^2(e^2 - 1) = (ae)^2 - a^2 = 25 - 16 = 9$$

$$\text{Equation of hyperbola: } \frac{x^2}{16} - \frac{y^2}{9} = 1$$

Testing point $$(8, 3\sqrt{3})$$ from List-II: $$\frac{64}{16} - \frac{27}{9} = 4 - 3 = 1 \implies \text{(S)} \rightarrow \text{(5)}$$