Tangent Line 06B693

1. **State the problem:** Determine if the line $\overline{QR}$ with equation $5y = -4x + 41$ is tangent to Circle $O$ at point $R(4,5)$. 2. **Rewrite the line equation in slope-intercept form:** $$5y = -4x + 41 \implies y = \frac{-4}{5}x + \frac{41}{5}$$ So, the slope of $\overline{QR}$ is $m_{QR} = -\frac{4}{5}$. 3. **Find the slope of radius $\overline{OR}$:** Since Circle $O$ is centered at the origin $(0,0)$ and $R$ is at $(4,5)$, the slope of $\overline{OR}$ is $$m_{OR} = \frac{5 - 0}{4 - 0} = \frac{5}{4}.$$ 4. **Check the product of slopes:** For $\overline{QR}$ to be tangent to the circle at $R$, the line must be perpendicular to the radius $\overline{OR}$. Two lines are perpendicular if the product of their slopes is $-1$. Calculate: $$m_{OR} \times m_{QR} = \frac{5}{4} \times \left(-\frac{4}{5}\right) = -1.$$ 5. **Conclusion:** Since the product of the slopes equals $-1$, $\overline{QR}$ is perpendicular to $\overline{OR}$ at $R$, so $\overline{QR}$ is tangent to Circle $O$ at $R$.