Is my solution correct?
Construction : Join $PS$ and $OR$.
$\angle ORQ = 90^o$ (Radius is perpendicular to tangent at the point of contact)
In $\triangle ORQ$,
$ OQ^2 = OR^2 + RQ^2 $ (using Pythagoras Theorem)
$\implies 13^2 = 8^2 + RQ^2$
$\implies RQ = \sqrt{105}$ $...(\mathtt i)$
Now, $SR=RQ$ (perpendicular from the center of the circle to a chord, bisects the chord)
$\implies SR = \sqrt{105}$ $...(\mathtt{ii})$
Adding $(\mathtt i)$ and $(\mathtt{ii})$,
$SR+RQ=2 \sqrt{105}$
$\implies SQ =2 \sqrt{105} $ $...(\mathtt{iii})$
Now, $\angle PSQ= 90^o$ (angle subtended by a diameter)
In $\triangle PSQ$ ,
$PQ^2=PS^2+SQ^2$ (using Pythagoras Theorem)
$\implies 26^2 = PS^2 + (2\sqrt{105})^2$ ($\because SQ= 2\sqrt{105}$, using $(\mathtt{iii})$ )
$\implies 676 = PS^2 + 420$
$\implies PS^2 = 676-420$
$\implies PS = \sqrt{256}$
$\implies PS = 16$ $...(\mathtt{iv})$
Now, In $\triangle PSR$,
$PR^2 = PS^2 + SR^2$ (using Pythagoras Theorem)
$\implies PR^2 = 16^2 + (\sqrt{105})^2$ ($\because SR= \sqrt{105}$, using $(\mathtt{ii})$ )
$\implies PR^2 = 256 + 105$
$\implies PR^2 = 361$
$\implies PR = \sqrt{361}$
$\implies PR = 19 cm$
Forgetting for a time the possible options : just be lazy if they really want an answer; you have a small number of data points $n$; so, fit a polynomial of degree $(n-1)$ which will go through all the $n$ points.
In the case you gave $$y=-\frac{x^3}{9}+\frac{22 x^2}{9}-\frac{47 x}{3}+36$$ So, for any new value of $x$, you will get a $y$.
For sure, the problem could be more complicated if $y$ must be an integer. So $(0,36)$, $(12,8)$, $(15,-24)$, $(16,-44)$, $(18,-102)$. If they expect $x$ and $y$ to be positive, plot the function : it is always negative if $x>14$. Since you do not repeat an ordered pair that is in the table, this let you a very limited choice.
Check your list of options.
Best Answer
The answer you state as "correct" is wrong. You check this through unit analysis: If radius and height are given in meters, the expression $r+1$ adds meters to a pure number, which is meaningless. Since you want area, the unit should become meters squared.
Your next-to-last line was correct: just finish the simplification for the really correct answer.
$$f(r)=2\pi r^2+6\pi r^2$$ $$=8\pi r^2$$
You can see that this has the right dimensions.