A curve has the property that the normal line through any point on the curve passes through $(2,0)$. Find its equation. A curve has the property that the normal line through any point on the curve passes through $(2,0)$.  If the curve contains the point $(2,3)$ find its equation.
My attempt:
Assume $(a,b)$ is a point on the function. The normal line has a slope $m=\frac{b}{a-2}$ and hence the tangent line has a slope $m_T = \frac{2-a}{b}$. Hence, this is the slope we can integrate to get the equation:
$$F(x) = \int \frac{2-a}{b}x dx = \frac{2-a}{2b}x^2+C$$
Since we know that $(2,3)$ is a point we can isolate for $C$ to get:
$$F(x) = \frac{2-a}{2b}x^2+\frac{3b-4+2a}{b}$$
I am getting stuck here because I am not sure how to get values for a $a$ and $b$...
 A: A circle has the property that the normal through any point on the curve will pass through its center.  As such, we can deduce that the center of the circle is $(2,0)$ and the circle contains the point $(2,3)$.  Thus, we have the equation $$(x-2)^2+y^2=9.$$

Taking an approach using calculus instead of geometry:
Define $-\dfrac{dx}{dy}$ as the slope of the normal to the curve $y(x)$.  The equation of the normal to the curve that contains $(x_1,y_1)$ would be $$y-y_1=-\frac{dx}{dy}(x-x_1).$$
Since the normal must contain $(2,0)$, we can substitute and solve a differential equation:
\begin{align}
y-0&=-\frac{dx}{dy}(x-2)\\
y\,dy&=(2-x)\,dx\\
\int y\,dy&=\int2-x\,dx\\
\frac12y^2&=-\frac12(2-x)^2+C_1\\
(x-2)^2+y^2&=C_2
\end{align}
Since this curve must contain $(2,3)$, it follows that $C_2=9$.  Therefore, the curve must have the equation $$(x-2)^2+y^2=9.$$
A: A somewhat more general proof, but one which is nevertheless in the spirit of our colleague Andrew Chin's answer, may be had using elementary differential geometry:
A Little Lemma:  Let $\alpha(s)$ be an arc-length parametrized curve in $\Bbb R^2$ such that every normal line to $\alpha(s)$ passes through the point $C \in \Bbb R^2$.  Then $\alpha(s)$ lies in a circle centered at $C$.
Proof of Little Lemma: Let $T(s)$ be the unit tangent vector field to $\alpha(s)$; that is
$T(s) = \dot \alpha(s); \tag 1$
by the first Frenet-Serret equation, $T$ is related to the curvature $\kappa(s)$ of $\alpha(s)$ via the formula
$\dot T(s) = \kappa(s) N(s), \tag 2$
where
$\kappa(s) > 0 \tag 3$
and $N(s)$ is the unit normal vector field to $\alpha(s)$.  Then the line through
$\alpha(s)$ in the direction $N(s)$ passes through $C$; if $l(s)$ denotes the signed length of the segment joining $\alpha(s)$ and $C$, we have
$\alpha(s) + l(s)N(s) = C, \tag 4$
holding for all $s$ in the domain of $\alpha(s)$.  (Note that $l(s)$ is differentiable in $s$ since we may write (4) in the form
$l(s)N(s) = C - \alpha(s), \tag{4.1}$
from which
$l(s) = l(s)N(s) \cdot N(s) = (C - \alpha(s)) \cdot N(s), \tag{4.2}$
with the right-hand side differentiable in $s$.)
We differentiate equation (4) with respect to $s$, and obtain
$\dot \alpha(s) + \dot l(s) N(s) + l(s) \dot N(s) = 0, \tag 5$
since $C$ does not depend on $s$.
The companion equation to (2),
$\dot N(s) = -\kappa(s)T(s), \tag 6$
which may be derived starting from the orthogonality of $T$ and $N$,
$T(s) \cdot N(s) = 0, \tag 7$
and then differentiating:
$\dot T(s) \cdot N(s) + T(s) \cdot \dot N(s) = 0, \tag 8$
and substituting $\dot T(s)$ from (2):
$\kappa(s) N(s) \cdot N(s) + T(s) \cdot \dot N(s) = 0; \tag 9$
since
$N(s) \cdot N(s) = 1, \tag{10}$
(9) may be written
$T(s) \cdot \dot N(s) = -\kappa(s); \tag{11}$
now, differentiating (10) yields
$\dot N(s) \cdot N(s) = 0, \tag{12}$
which shows that $\dot N(s)$ is parallel to $T(s)$; then in light of (11) we infer
$\dot N(s) = -\kappa(s)T(s), \tag{13}$
and substituting this and (1) into (5) we write
$T(s) + \dot l(s) N(s) - l(s) \kappa(s) T(s) = 0, \tag{14}$
i.e.,
$(1 - l(s) \kappa(s))T(s) + \dot l(s) N(s) = 0; \tag{15}$
the orthnormality of $T(s)$ and $N(s)$ now allows us to conclude that
$1 - l(s) \kappa(s) = 0, \tag{16}$
$\dot l(s) = 0, \tag{17}$
which show that both $l(s) = l$ and $\kappa(s) = \kappa$ are constant with
$\kappa = \dfrac{1}{l}, \tag{18}$
and hence in light of (3) that
$l > 0; \tag{18.1}$
finally, from (4)
$\vert \alpha(s) - C \vert = \vert l(s)N(s) \vert = l(s) \vert N(s) \vert = l, \tag{19}$
which shows that $\alpha(s)$ lies in the circle of radius $l$ centered at $C$.  End:  Proof of Litte Lemma.
Now taking
$C = (2, 0) \tag{20}$
and
$\alpha(s) = (2, 3) \tag{21}$
for some $s$, we find that
$l = 3, \tag{22}$
and thus see that $\alpha(s)$ lies in the circle of radius $3$ about $(2, 0)$, the same result as obtained by Andrew Chin in his answer.  The equation of the circle in which $\alpha(s)$ lies is of course
$(x - 2)^2 + y^2 = 9. \tag{23}$
