1. **Problem statement:** We are given three points $P(2, -4)$, $Q(-4, -4)$, and $R(-4, 6)$ that lie on a circle. We need to find the quadrant in which the center of this circle lies. 2. **Formula and concept:** The center of a circle passing through three points is the intersection of the perpendicular bisectors of the segments joining these points. 3. **Step 1: Find midpoints of segments $PQ$ and $QR$** - Midpoint of $PQ$: $$M_{PQ} = \left(\frac{2 + (-4)}{2}, \frac{-4 + (-4)}{2}\right) = (-1, -4)$$ - Midpoint of $QR$: $$M_{QR} = \left(\frac{-4 + (-4)}{2}, \frac{-4 + 6}{2}\right) = (-4, 1)$$ 4. **Step 2: Find slopes of $PQ$ and $QR$** - Slope of $PQ$: $$m_{PQ} = \frac{-4 - (-4)}{2 - (-4)} = \frac{0}{6} = 0$$ - Slope of $QR$: $$m_{QR} = \frac{6 - (-4)}{-4 - (-4)} = \frac{10}{0} = \text{undefined (vertical line)}$$ 5. **Step 3: Find slopes of perpendicular bisectors** - Perpendicular bisector of $PQ$ has slope $$m_{\perp PQ} = -\frac{1}{m_{PQ}} = -\frac{1}{0} = \text{undefined (vertical line)}$$ - Perpendicular bisector of $QR$ has slope $$m_{\perp QR} = -\frac{1}{m_{QR}} = -\frac{1}{\text{undefined}} = 0$$ 6. **Step 4: Write equations of perpendicular bisectors** - Perpendicular bisector of $PQ$ passes through $M_{PQ}(-1, -4)$ and is vertical: $$x = -1$$ - Perpendicular bisector of $QR$ passes through $M_{QR}(-4, 1)$ and has slope 0 (horizontal line): $$y = 1$$ 7. **Step 5: Find intersection of perpendicular bisectors** - Intersection point is at $$x = -1$$ and $$y = 1$$ 8. **Step 6: Determine the quadrant of the center $C(-1, 1)$** - $x = -1$ (negative), $y = 1$ (positive) means the center lies in **Quadrant II**. **Final answer:** The center of the circle lies in Quadrant II.