ทุนญี่ปุ่น 2019B

[nowhere]

Consider the circle $x^{2}+y^{2}=r^{2} (r>0)$ and the line $y=m(x-5)+3$. The line intersects the circle if and only if $0\leqslant m\leqslant ?$. Also $r=?$.

[Anton]

Problem. Consider the circle $$x^2+y^2=r^2$$ and the line $$y=m(x-5)+3\,,$$ where $r$ is a positive real number and $m$ is a real number. It is known that the line intersects the circle if and only if $0\leq m\leq \mu$ for some real number $\mu$. Determine the values of $\mu$ and $r$.

Observe that $y=m(x-5)+3$ is an equation of a line passing through a fixed point $P(5,3)$. When $m=0$, this line is a horizontal line given by the equation $y=3$. From the information given by the problem statement, the minimum possible value of the slope $m$ is $0$, then the horizontal line $y=3$ is tangent to the circle. Therefore, we must have $r=3$.

Now, we have to calculate the possible range of $m$. Since $r=3$ has been established, we let $A=(0,3)$ be a point such that $PA$ is tahgnet to the circle $C$ given by the equation $x^2+y^2=r^2=3^2=9$. Suppose that $B\neq A$ is another point on the circle $C$ such that $PB$ is tangent to $C$. Suppose that $B=(u,v)$ for some coordinates $u$ and $v$. Then, we know that $AB\perp OP$, where $O$ is the origin, which is the center of $C$.

Since the equation of the line $OP$ is $y=\dfrac{3}{5}\,x$, the equation of the line $AB$ must be
$$y=-\frac{5}{3}\,x+3\,.$$
Therefore, $u^2+v^2=9$ because $B$ lies on $C$, as well as $v=-\frac{5}{3}\,u+3$, because $B$ lies on $AB$. This gives
$$u^2+\left(-\frac{5}{3}\,u+3\right)^2=9\,.$$
Hence,
$$u^2+\frac{25}{9}\,u^2-10\,u+9=9\,.$$
Consequently,
$$\frac{34}{9}\,u^2-10\,u=0\,.$$
Because $u\neq 0$ (otherwise $v=3$, so $B=A$, which contradicts the assumption that $B\neq A$), we get
$$u=\frac{45}{17}\,.$$
Therefore,
$$v=-\frac{5}{3}\,u+3=-\frac{75}{17}+3=-\frac{24}{17}\,.$$
The maximum value of $m$ is thus
$$\frac{3-v}{5-u}=\frac{3+\frac{24}{17}}{5-\frac{45}{17}}=\frac{51+24}{85-45}=\frac{75}{40}=\frac{15}{8}\,.$$