4. Given points $A(1,0)$ and $B(4,3)$. For a real number $p$, denote by $N_{p}$ the point on the line $AB$ that is closest to the circle

$$
(x-p)^{2}+\left(y-1-p^{2}\right)^{2}=1
$$

(under the distance of a point from a circle, we mean the distance from that point to the nearest point on the circle). For which values of the parameter $p$ is the point $N_{p}$ strictly between points $A$ and $B$?
4. The equation of the line passing through points $A$ and $B$ is $y=x-1$. The circle in question has its center at the point $\left(p, p^{2}+1\right)$ and a radius of 1. We will prove that this circle does not intersect the line $A B$. For this purpose, we will first determine the normal through the point $\left(p, p^{2}+1\right)$ to the line $A B$. Its slope is $-\frac{1}{1}=-1$, so since its equation is of the form $y=-x+n$, the value of $n$ is found from the condition that the drawn normal passes through the point $\left(p, p^{2}+1\right)$: we have $p^{2}+1=-p+n$, so the equation of the considered normal is $y=-x+p^{2}+p+1$. The intersection of this normal with the line

![](https://cdn.mathpix.com/cropped/2024_06_04_f254ba4f25ce7f8d9df1g-61.jpg?height=385&width=442&top_left_y=1711&top_left_x=1074)

Dr 2017 3B 4
$A B$ (denote it as $\left(x_{0}, y_{0}\right)$) is found by substituting $y=x-1$ into the previous equation: it then becomes $x-1=-x+p^{2}+p+1$, from which we get $x_{0}=\frac{p^{2}+p+2}{2}$, and then $y_{0}=x_{0}-1=\frac{p^{2}+p}{2}$. This point on the line $A B$ is, therefore, the closest to the point $\left(p, p^{2}+1\right)$, so to prove that the observed line and circle do not intersect, it is sufficient to verify that the distance from the center of the circle to this point is greater than the radius. We calculate the square of this distance:

$$
\begin{aligned}
\left(x_{0}-p\right)^{2}+\left(y_{0}-\left(p^{2}+1\right)\right)^{2} & =\left(\frac{p^{2}-p+2}{2}\right)^{2}+\left(\frac{-p^{2}+p-2}{2}\right)^{2}=2\left(\frac{p^{2}-p+2}{2}\right)^{2} \\
& =\frac{\left(p^{2}-p+2\right)^{2}}{2}=\frac{\left(\left(p-\frac{1}{2}\right)^{2}+\frac{7}{4}\right)^{2}}{2} \geqslant \frac{\left(\frac{7}{4}\right)^{2}}{2}=\frac{49}{32}>1
\end{aligned}
$$

which is what we needed to prove.

Since the line $A B$ and the circle do not intersect, the point on the line $A B$ closest to the circle is precisely the foot of the normal from the center of the circle to the line $A B$, which is the point $\left(x_{0}, y_{0}\right)$. For this point to lie between points $A$ and $B$, it must satisfy $10$; the right inequality reduces to $p^{2}+p-6<0$, i.e., $(p+3)(p-2)<0$, which is satisfied for $-3<p<2$. Combining the obtained constraints, we conclude that the solution to the problem is: $p \in(-3,-1) \cup(0,2)$.