Problem 233

To solve for the value of $m$ that satisfies the condition "$\triangle ABC$ has an area of $\frac{8}{5}$," we proceed as follows:

First Approach:

1. From the circle's equation $C: (x-1)^{2}+y^{2}=4$, we identify the center $C(1,0)$ and radius $r=2$.
2. Knowing the area of $\triangle ABC$ is $\frac{8}{5}$, we use the formula for the area of a triangle formed by two radii and a chord on a circle: $S_{\triangle ABC}=\frac{1}{2}\times 2\times 2\times \sin \angle ACB=\frac{8}{5}$.
3. Solving for $\sin \angle ACB$, we find $\sin \angle ACB=\frac{4}{5}$.
4. Letting $\frac{1}{2}\angle ACB=\theta$, we have $2\sin \theta \cos \theta =\frac{4}{5}$.
5. This simplifies to $\frac{2\tan\theta}{\tan^{2}\theta+1}=\frac{4}{5}$, leading to $\tan \theta =\frac{1}{2}$ or $\tan \theta =2$.
6. Therefore, $\cos \theta =\frac{2}{\sqrt{5}}$ or $\cos \theta =\frac{1}{\sqrt{5}}$.
7. The distance $d$ from the center to the line $x-my+1=0$ is either $d=\frac{4}{\sqrt{5}}$ or $d=\frac{2}{\sqrt{5}}$.
8. Solving $\frac{2}{\sqrt{1+m^{2}}}=\frac{4}{\sqrt{5}}$ or $\frac{2}{\sqrt{1+m^{2}}}=\frac{2}{\sqrt{5}}$, we find $m=\pm \frac{1}{2}$ or $m=\pm 2$.

Second Approach:

1. With the radius of circle $C$ being $2$ and center $C(1,0)$, we let the distance from center $C$ to the line $x-my+1=0$ be $d$.
2. The chord length $|AB|=2\sqrt{4-d^{2}}$, so $S_{\triangle ABC}=\frac{1}{2}\times 2\sqrt{4-d^{2}}\times d=\frac{8}{5}$.
3. Solving for $d^{2}$, we get $d^{2}=\frac{4}{5}$ or $d^{2}=\frac{16}{5}$, leading to $d=\frac{2\sqrt{5}}{5}$ or $d=\frac{4\sqrt{5}}{5}$.
4. Using the distance formula from a point to a line, when $d=\frac{2\sqrt{5}}{5}$, we get $\frac{2}{\sqrt{m^{2}+1}}=\frac{2\sqrt{5}}{5}$, resulting in $m=\pm 2$.
5. When $d=\frac{4\sqrt{5}}{5}$, the distance formula yields $\frac{2}{\sqrt{m^{2}+1}}=\frac{4\sqrt{5}}{5}$, leading to $m=\pm \frac{1}{2}$.

In conclusion, the possible values of $m$ are $\pm 2$ or $\pm \frac{1}{2}$. Therefore, the answer is $\boxed{2}$ (or $\boxed{-2}$ or $\boxed{\frac{1}{2}}$ or $\boxed{-\frac{1}{2}}$).

To solve for the value of $m$ that satisfies the condition "$\triangle ABC$ has an area of $\frac{8}{5}$," we proceed as follows:

First Approach:

1. Given the circle $C: \left(x-1\right)^{2}+y^{2}=4$, we identify its center as $C(1,0)$ and its radius as $r=2$.
2. The area of $\triangle ABC$, $S_{\triangle ABC}$, is given by $\frac{1}{2} \times 2 \times 2 \times \sin \angle ACB = \frac{8}{5}$. Simplifying, we find $\sin \angle ACB = \frac{4}{5}$.
3. Letting $\frac{1}{2}\angle ACB = \theta$, we have $2\sin \theta \cos \theta = \frac{4}{5}$. This implies $\frac{2\sin\theta\cos\theta}{\sin^{2}\theta+\cos^{2}\theta}=\frac{4}{5}$.
4. Simplifying further, we get $\frac{2\tan\theta}{\tan^{2}\theta+1}=\frac{4}{5}$. Solving for $\tan \theta$, we find $\tan \theta =\frac{1}{2}$ or $\tan \theta =2$.
5. This leads to $\cos \theta =\frac{2}{\sqrt{5}}$ or $\cos \theta =\frac{1}{\sqrt{5}}$.
6. The distance from the center to the line $x-my+1=0$ is $d=\frac{4}{\sqrt{5}}$ or $\frac{2}{\sqrt{5}}$. This gives us two equations: $\frac{2}{\sqrt{1+m^{2}}}=\frac{4}{\sqrt{5}}$ or $\frac{2}{\sqrt{1+m^{2}}}=\frac{2}{\sqrt{5}}$.
7. Solving these equations for $m$, we find $m=\pm \frac{1}{2}$ or $m=\pm 2$.

Second Approach:

1. With the radius of circle $C$ being $2$ and its center at $C(1,0)$, we let the distance from the center $C$ to the line $x-my+1=0$ be $d$.
2. The chord length $|AB|$ is given by $2\sqrt{4-d^{2}}$. Thus, $S_{\triangle ABC}=\frac{1}{2}\times 2\sqrt{4-d^{2}}\times d=\frac{8}{5}$.
3. Solving for $d^{2}$, we find $d^{2}=\frac{4}{5}$ or $d^{2}=\frac{16}{5}$. This implies $d=\frac{2\sqrt{5}}{5}$ or $d=\frac{4\sqrt{5}}{5}$.
4. When $d=\frac{2\sqrt{5}}{5}$, using the formula for the distance from a point to a line, we get $\frac{2}{\sqrt{m^{2}+1}}=\frac{2\sqrt{5}}{5}$, leading to $m=\pm 2$.
5. When $d=\frac{4\sqrt{5}}{5}$, we have $\frac{2}{\sqrt{m^{2}+1}}=\frac{4\sqrt{5}}{5}$, resulting in $m=\pm \frac{1}{2}$.

Therefore, combining the results from both approaches, we conclude that the possible values for $m$ are $\pm 2$ or $\pm \frac{1}{2}$. Thus, the answer is $\boxed{2}$ (or $-2$ or $\frac{1}{2}$ or $-\frac{1}{2}$).