# **Step-by-Step Solution** --- ## **Part A: Deriving the Standard Equation of a Circle** **Using the Pythagorean Theorem:** - The distance from the center \((r, s)\) to a point on the circle \((x, y)\) is the radius \(t\). - By the distance formula (from Pythagorean Theorem): \[ \text{Distance} = \sqrt{(x - r)^2 + (y - s)^2} \] - For any point \((x, y)\) on the circle, this equals the radius \(t\): \[ \sqrt{(x - r)^2 + (y - s)^2} = t \] - Square both sides to get the **standard equation of a circle**: \[ (x - r)^2 + (y - s)^2 = t^2 \] --- ## **Part B: Domain and Range of the Circle** Given: - Center: \((r, s) = (7, -4)\) - Radius: \(t = 10\) **Equation:** \((x - 7)^2 + (y + 4)^2 = 100\) ### **Domain (possible \(x\)-values):** - The furthest left: \(r - t = 7 - 10 = -3\) - The furthest right: \(r + t = 7 + 10 = 17\) **Domain:** \[ \boxed{[-3, 17]} \] ### **Range (possible \(y\)-values):** - The lowest: \(s - t = -4 - 10 = -14\) - The highest: \(s + t = -4 + 10 = 6\) **Range:** \[ \boxed{[-14, 6]} \] --- ## **Part C: Is the Point (9, 1) Inside the Circle?** **Given:** - Center: \((7, -4)\) - Radius: \(t = 10\) - Point: \((9, 1)\) **Plug into the circle equation:** \[ (x - r)^2 + (y - s)^2 < t^2 \quad \text{(for inside)} \] Plug values: \[ (9 - 7)^2 + (1 - (-4))^2 = (2)^2 + (5)^2 = 4 + 25 = 29 \] Compare with \(t^2 = 100\): \[ 29 < 100 \] **Conclusion:** Since \(29 < 100\), the point \((9, 1)\) is **inside the border of the circle**. --- ## **Final Answers** ### **Part A:** \[ \boxed{(x - r)^2 + (y - s)^2 = t^2} \] ### **Part B:** \[ \text{Domain: } [-3, 17] \quad \text{Range: } [-14, 6] \] ### **Part C:** \[ \text{Yes, the point } (9, 1) \text{ is inside the circle because } 29 < 100. \]