I4.4 Let $x$ and $y$ be real numbers satisfying the equation $(x-\sqrt{p})^{2}+(y-\sqrt{p})^{2}=1$.
If $k=\frac{y}{x-3}$ and $q$ is the least possible values of $k^{2}$, find the value of $q$.
$$
(x-3)^{2}+(y-3)^{2}=1 \ldots
$$

Sub. $y=k(x-3)$ into (1): $(x-3)^{2}+[k(x-3)-3]^{2}=1$
$$
\begin{array}{l}
(x-3)^{2}+k^{2}(x-3)^{2}-6 k(x-3)+9=1 \\
\left(1+k^{2}\right)(x-3)^{2}-6 k(x-3)+8=0 \Rightarrow\left(1+k^{2}\right) t^{2}-6 k t+8=0 ; \text { where } t=x-3
\end{array}
$$

For real values of $t: \Delta=4\left[3^{2} k^{2}-8\left(1+k^{2}\right)\right] \geq 0$
$$
k^{2} \geq 8
$$

The least possible value of $k^{2}=q=8$.
Method 2
The circle $(x-3)^{2}+(y-3)^{2}=1$ intersects with the variable line $y=k(x-3)$ which passes through a fixed point $(3,0)$, where $k$ is the slope of the line.
From the graph, the extreme points when the variable line touches the circle at $B$ and $C$. $H(3,3)$ is the centre of the circle.
In $\triangle A B H, \angle A B H=90^{\circ}$ ( $\angle$ in semi-circle $)$
Let $\angle H A B=\theta, A H=3, H B=1$,
$A B=\sqrt{8}=2 \sqrt{2} \quad$ (Pythagoras' theorem)
$$
\tan \theta=\frac{1}{\sqrt{8}}
$$

Slope of $A B=\tan \left(90^{\circ}-\theta\right)=\sqrt{8}$
Least possible $k^{2}=q=8$