5Question: Find the area of the region enclosed by the circle \(x^2 + y^2 = 16\) and the line \(y = x + 2\).

["Title: How to Find the Area of the Region Enclosed by the Circle (x^2 + y^2 = 16) and the Line (y = x + 2) – A Step-by-Step Guide", "If you're looking to determine the area of the region enclosed by a circle and a line, you're tackling a classic problem in coordinate geometry and calculus. In this article, we’ll explore how to find the area bounded by the circle (x^2 + y^2 = 16) and the line (y = x + 2) step-by-step. This exercise combines algebraic analysis with geometric reasoning—essential skills for mastering areas in coordinate systems.", "---", "### Understanding the Problem", "We are given:", "- A circle defined by (x^2 + y^2 = 16), which has a center at the origin ((0, 0)) and radius (4).\n- A straight line defined by (y = x + 2).\nWe need to find the area of the region shared by both—that is, the finite region bounded by the circle arc and the line segment where they intersect.", "This area can be found by computing the segment of the circle cut out by the line, combining circle geometry and integration.", "---", "### Step 1: Find Points of Intersection", "To locate where the line intersects the circle, substitute (y = x + 2) into the circle’s equation:", "[\nx^2 + (x + 2)^2 = 16\n]", "Expand and simplify:", "[\nx^2 + (x^2 + 4x + 4) = 16 \Rightarrow 2x^2 + 4x + 4 = 16 \Rightarrow 2x^2 + 4x - 12 = 0\n]", "Divide through by 2:", "[\nx^2 + 2x - 6 = 0\n]", "Use the quadratic formula:", "[\nx = \frac{-2 \pm \sqrt{2^2 - 4(1)(-6)}}{2} = \frac{-2 \pm \sqrt{4 + 24}}{2} = \frac{-2 \pm \sqrt{28}}{2} = \frac{-2 \pm 2\sqrt{7}}{2} = -1 \pm \sqrt{7}\n]", "So the (x)-coordinates of intersection are (x = -1 - \sqrt{7}) and (x = -1 + \sqrt{7}).", "Now find the corresponding (y)-values using (y = x + 2):", "- For (x = -1 - \sqrt{7}), (y = 1 - \sqrt{7})\n- For (x = -1 + \sqrt{7}), (y = 1 + \sqrt{7})", "Thus, the intersection points are:\n(A = (-1 - \sqrt{7},\ 1 - \sqrt{7}))\n(B = (-1 + \sqrt{7},\ 1 + \sqrt{7}))", "---", "### Step 2: Determine the Central Angle Subtended by the Chord", "The enclosed region is a circular segment bounded by the arc from (A) to (B) and the chord (AB). To compute the area, we need the central angle (\ heta) (in radians) subtended by chord (AB) at the origin.", "Use the dot product formula with vectors (\vec{OA}) and (\vec{OB}):", "Let\n( \vec{OA} = \langle -1 - \sqrt{7},\ 1 - \sqrt{7} \rangle )\n( \vec{OB} = \langle -1 + \sqrt{7},\ 1 + \sqrt{7} \rangle )", "Compute the dot product:", "[\n\vec{OA} \cdot \vec{OB} = (-1 - \sqrt{7})(-1 + \sqrt{7}) + (1 - \sqrt{7})(1 + \sqrt{7})\n]", "Calculate each term:", "- First: ( (-1)^2 - (\sqrt{7})^2 = 1 - 7 = -6 )\n- Second: (1^2 - (\sqrt{7})^2 = 1 - 7 = -6)\nSo, dot product = (-6 + (-6) = -12)", "Now magnitudes:", "[\n|\vec{OA}| = \sqrt{(-1 - \sqrt{7})^2 + (1 - \sqrt{7})^2} = \sqrt{(1 + 2\sqrt{7} + 7) + (1 - 2\sqrt{7} + 7)} = \sqrt{16 + 8} = \sqrt{24} = 2\sqrt{6}\n]", "Similarly, (|\vec{OB}| = 2\sqrt{6})", "So,", "[\n\cos\ heta = \frac{\vec{OA} \cdot \vec{OB}}{|\vec{OA}||\vec{OB}|} = \frac{-12}{(2\sqrt{6})(2\sqrt{6})} = \frac{-12}{24} = -\frac{1}{2}\n]", "Thus, (\ heta = \frac{2\pi}{3}) radians (since cosine is (-\frac{1}{2}) and angle is between 0 and (\pi)).", "---", "### Step 3: Compute the Area of Circular Segment", "The area of the region (segment) bounded by the chord and the arc is:", "[\n\ ext{Segment Area} = \ ext{Area of Sector} - \ ext{Area of Triangle }\n]", "- Area of Sector:\n[\n\frac{1}{2} r^2 \ heta = \frac{1}{2} \cdot 16 \cdot \frac{2\pi}{3} = \frac{16\pi}{3}\n]", "- Area of Triangle OAB: Use half the magnitude of the cross product of vectors (\vec{OA}) and (\vec{OB}):", "[\n|\vec{OA} \ imes \vec{OB}| = \left| x_1 y_2 - x_2 y_1 \right| = \left| (-1 - \sqrt{7})(1 + \sqrt{7}) - (-1 + \sqrt{7})(1 - \sqrt{7}) \right|\n]", "Calculate:", "- First term: ((-1)(1) + (-1)(\sqrt{7}) + (-\sqrt{7})(1) + (-\sqrt{7})(\sqrt{7}) = -1 - \sqrt{7} - \sqrt{7} - 7 = -8 - 2\sqrt{7})\n- Second term: ((-1)(1) + (-1)(-\sqrt{7}) + (\sqrt{7})(1) + (\sqrt{7})(-\sqrt{7}) = -1 + \sqrt{7} + \sqrt{7} - 7 = -8 + 2\sqrt{7})\nSo difference: ((-8 - 2\sqrt{7}) - (-8 + 2\sqrt{7}) = -4\sqrt{7}) → absolute value: (4\sqrt{7})", "Thus, triangle area:", "[\n\frac{1}{2} \cdot \frac{1}{2} | \vec{OA} \ imes \vec{OB} | = \frac{1}{4} \cdot 4\sqrt{7} = \sqrt{7}\n]", "Wait — correction: The cross product magnitude is (| -4\sqrt{7} | = 4\sqrt{7}), so area of triangle is:", "[\n\frac{1}{2} \cdot \ ext{base} \cdot \ ext{height} \quad \ ext{or directly:} \quad \frac{1}{2} |\vec{OA} \ imes \vec{OB}| = \frac{1}{2} \cdot 4\sqrt{7} = 2\sqrt{7}\n]", "Actually, correction: The formula for area from cross product is:", "[\n\ ext{Area} = \frac{1}{2} |x_1 y_2 - x_2 y_1| = \frac{1}{2} |4\sqrt{7}| = 2\sqrt{7}\n]", "So:", "[\n\ ext{Segment Area} = \frac{16\pi}{3} - 2\sqrt{7}\n]", "This is positive since (\frac{16\pi}{3} \approx 16.755), (2\sqrt{7} \approx 5.291), so (16.755 - 5.291 > 0).", "---", "### Final Area of Enclosed Region", "Since the line cuts through the circle and forms a single bounded region above the chord, the enclosed area is exactly the area of the circular segment:", "[\n\boxed{\frac{16\pi}{3} - 2\sqrt{7}}\n]", "This represents the finite area enclosed by the circle (x^2 + y^2 = 16) and the line (y = x + 2), computed by combining sector and triangle areas with precise intersection geometry.", "---", "### Why This Matters and How to Use This Knowledge", "Understanding how to derive such areas helps in fields ranging from engineering design to physics and computer graphics, where computing enclosed regions in coordinate systems is critical. Mastering geometric intersection, angular computations, and integral-based area methods builds a strong foundation for advanced calculus and 3D modeling.", "---", "Keywords: area enclosed by circle and line, intersecting circle and line area, find area bounded by circle and line, geometry of circle segments, step-by-step circle integration, solve area between line and circle", "Meta Description: Learn how to compute the area enclosed by the circle (x^2 + y^2 = 16) and line (y = x + 2) using intersection points, central angle, sector, and triangle area formulas. Step-by-step guide for students and math enthusiasts.", "---", "Use this method for similar problems by: identifying curves, solving for intersections, computing central angle via dot product, then applying segment area formula."]