{
  "format": "stapler-pack",
  "version": 1,
  "kind": "quiz",
  "meta": {
    "title": "Circles, Parabolas & Constructions",
    "author": "Staples Education",
    "description": "Circles in standard form and by completing the square, transcribed from the tutoring whiteboard archive. Parabolas and compass-and-straightedge constructions are authored to complete these shelves; the archive does not reach them.",
    "created": "2026-08-06T04:14:56.059Z",
    "license": "Proprietary - Staples Education"
  },
  "payload": {
    "quiz": {
      "id": "coordinate-conics",
      "title": "Circles, Parabolas & Constructions",
      "blurb": "Standard form and completing the square for circles, the focus-directrix definition of a parabola, and the compass-and-straightedge constructions behind both.",
      "meta": "19 questions · easy → hard",
      "categories": [
        "circles-coordinate-plane",
        "parabolas-coordinate-plane",
        "constructions-circles"
      ],
      "questions": [
        {
          "id": "ccp-001",
          "topic": "Circles on the Coordinate Plane",
          "difficulty": "medium",
          "stem": "What are the centre and radius of the circle (x − 3)² + (y + 1)² = 25?",
          "choices": [
            "centre (3, −1), radius 5",
            "centre (−3, 1), radius 5",
            "centre (3, −1), radius 25",
            "centre (3, 1), radius 5"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. Standard form is (x − h)² + (y − k)² = r², so the centre is (h, k) = (3, −1) — note that (y + 1) is (y − (−1)) — and r = √25 = 5.",
            "This flips both signs. The form SUBTRACTS the coordinates, so (x − 3) gives h = +3, not −3.",
            "25 is r², not r. The square root is the last step.",
            "The y-coordinate sign is wrong. (y + 1) means k = −1, because +1 is what you get by subtracting −1."
          ],
          "hint": "Standard form subtracts. Rewrite (y + 1) as (y − ?) and read the centre straight off."
        },
        {
          "id": "ccp-002",
          "topic": "Circles on the Coordinate Plane",
          "difficulty": "hard",
          "stem": "Complete the square to find the radius of the circle x² + y² + 4x − 6y − 12 = 0.",
          "choices": [
            "5",
            "25",
            "12",
            "√12"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. Group and complete: (x² + 4x + 4) + (y² − 6y + 9) = 12 + 4 + 9, giving (x + 2)² + (y − 3)² = 25. So r² = 25 and r = 5, centred at (−2, 3).",
            "25 is r². One square root short.",
            "12 is the constant that was moved across before the completing terms were added. Both 4 and 9 still have to be added to it.",
            "√12 uses the constant before completing the square. The 4 and the 9 added to the left must also be added to the right."
          ],
          "hint": "Half the x-coefficient squared, and half the y-coefficient squared. Add both to BOTH sides."
        },
        {
          "id": "ccp-003",
          "topic": "Circles on the Coordinate Plane",
          "difficulty": "medium",
          "stem": "Write the equation of the circle with centre (2, −5) and radius 3.",
          "choices": [
            "(x − 2)² + (y + 5)² = 9",
            "(x + 2)² + (y − 5)² = 9",
            "(x − 2)² + (y + 5)² = 3",
            "(x − 2)² + (y − 5)² = 9"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. Substituting h = 2, k = −5, r = 3 into (x − h)² + (y − k)² = r² gives (x − 2)² + (y − (−5))² = 3², i.e. (x − 2)² + (y + 5)² = 9.",
            "Both signs are flipped. The form subtracts the centre's coordinates, so a centre of +2 produces (x − 2).",
            "The right-hand side must be r², which is 9, not r.",
            "The y term has the wrong sign. Subtracting −5 gives +5 inside the bracket."
          ],
          "hint": "Substitute into the standard form carefully, remembering that subtracting a negative flips the sign."
        },
        {
          "id": "ccp-004",
          "topic": "Circles on the Coordinate Plane",
          "difficulty": "medium",
          "stem": "A circle has a diameter with endpoints (1, 2) and (7, 10). What is its centre and radius?",
          "choices": [
            "centre (4, 6), radius 5",
            "centre (4, 6), radius 10",
            "centre (3, 4), radius 5",
            "centre (8, 12), radius 5"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. The centre is the midpoint of the diameter: ((1+7)/2, (2+10)/2) = (4, 6). The diameter's length is √(6² + 8²) = 10, so the radius is 5.",
            "10 is the DIAMETER, not the radius. Halve it.",
            "(3, 4) is the vector from one endpoint to the centre, or half the difference — not the midpoint itself.",
            "(8, 12) is the sum of the coordinates without halving."
          ],
          "hint": "The centre is the midpoint. The radius is half the distance between the endpoints — do not stop at the diameter."
        },
        {
          "id": "ccp-005",
          "topic": "Parabolas",
          "difficulty": "medium",
          "stem": "Geometrically, a parabola is the set of all points that are:",
          "choices": [
            "equidistant from a fixed point (the focus) and a fixed line (the directrix)",
            "equidistant from two fixed points",
            "a fixed distance from a single fixed point",
            "the sum of whose distances from two fixed points is constant"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. That is the defining property. Every point on a parabola is exactly as far from the focus as it is from the directrix, and the vertex is the point halfway between them.",
            "Points equidistant from two fixed points form the PERPENDICULAR BISECTOR of the segment joining them — a straight line, not a curve.",
            "A fixed distance from a single point describes a CIRCLE.",
            "A constant SUM of distances from two fixed points describes an ELLIPSE."
          ],
          "hint": "Each of the four descriptions defines a different figure. Which one involves a point and a line rather than two points?",
          "diagram": "<svg viewBox=\"0 0 240 152\" width=\"100%\" style=\"max-width:240px; height:auto; display:block; margin:0 auto;\" role=\"img\"><path d=\"M 40,20 Q 120,150 200,20\" fill=\"none\" stroke=\"#7c3aed\" stroke-width=\"2\"/><line x1=\"30\" y1=\"128\" x2=\"210\" y2=\"128\" stroke=\"#6b6b76\" stroke-width=\"1.6\" stroke-linecap=\"round\"/><circle cx=\"120\" cy=\"62\" r=\"3.2\" fill=\"#c0392b\"/><circle cx=\"120\" cy=\"95\" r=\"3.2\" fill=\"#1f2024\"/><circle cx=\"70\" cy=\"51\" r=\"3.2\" fill=\"#7c3aed\"/><line x1=\"70\" y1=\"51\" x2=\"120\" y2=\"62\" stroke=\"#c0392b\" stroke-width=\"1.5\" stroke-dasharray=\"5 4\" stroke-linecap=\"round\"/><line x1=\"70\" y1=\"51\" x2=\"70\" y2=\"128\" stroke=\"#c0392b\" stroke-width=\"1.5\" stroke-dasharray=\"5 4\" stroke-linecap=\"round\"/><text x=\"126\" y=\"58\" fill=\"#c0392b\" font-size=\"10.5\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">focus</text><text x=\"126\" y=\"100\" fill=\"#6b6b76\" font-size=\"10.5\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">vertex</text><text x=\"150\" y=\"142\" fill=\"#6b6b76\" font-size=\"10.5\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">directrix</text><text x=\"40\" y=\"44\" fill=\"#1f2024\" font-size=\"11\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">P</text></svg>",
          "diagramCaption": "The two dashed distances from P are always equal. Not drawn to scale"
        },
        {
          "id": "ccp-006",
          "topic": "Parabolas",
          "difficulty": "hard",
          "stem": "A parabola has its vertex at the origin and its focus at (0, 2). What is its equation?",
          "choices": [
            "x² = 8y",
            "y² = 8x",
            "x² = 2y",
            "x² = 4y"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. For a vertical parabola with vertex at the origin the form is x² = 4py, where p is the directed distance from vertex to focus. Here p = 2, so x² = 8y. The directrix is the line y = −2.",
            "y² = 8x opens sideways, which happens when the focus is on the x-axis. This focus is on the y-axis, so the parabola opens upward.",
            "x² = 2y would put the focus at (0, ½), since 4p = 2 gives p = ½.",
            "x² = 4y would put the focus at (0, 1). The 4p must equal 8 when p = 2."
          ],
          "hint": "Use x² = 4py and be careful: p is the vertex-to-focus distance, and it gets multiplied by 4."
        },
        {
          "id": "ccp-007",
          "topic": "Parabolas",
          "difficulty": "medium",
          "stem": "For the parabola x² = 12y, where is the directrix?",
          "choices": [
            "y = −3",
            "y = 3",
            "x = −3",
            "y = −12"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. Comparing with x² = 4py gives 4p = 12, so p = 3. The focus is at (0, 3) and the directrix is the mirror line on the other side of the vertex: y = −3.",
            "y = 3 is the height of the FOCUS, not the directrix. They sit on opposite sides of the vertex.",
            "x = −3 is a vertical line, which would be the directrix of a sideways-opening parabola. This one opens upward.",
            "y = −12 uses 4p rather than p. Divide by 4 first."
          ],
          "hint": "Read 4p off the equation, halve it twice to get p, then place the directrix the same distance below the vertex as the focus is above."
        },
        {
          "id": "ccp-008",
          "topic": "Constructions",
          "difficulty": "medium",
          "stem": "To construct the perpendicular bisector of segment AB with compass and straightedge, what do you do first?",
          "choices": [
            "Draw arcs of equal radius (more than half of AB) centred at A and at B",
            "Measure AB with a ruler and mark the midpoint",
            "Draw a circle with AB as its diameter",
            "Draw arcs of any two different radii from A and B"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. Equal radii from both endpoints make the two intersection points equidistant from A and B. Since the perpendicular bisector IS the set of points equidistant from A and B, the line through those two intersections is it. The radius must exceed half of AB or the arcs never meet.",
            "Measuring is not a construction. Compass-and-straightedge work is valid precisely because it uses no marked ruler — the point is to produce the midpoint, not to measure it.",
            "A circle on AB as diameter passes through A and B, and its centre is already the midpoint you are trying to find — so this assumes the answer.",
            "Different radii would give points that are NOT equidistant from A and B, so the line through them would not be the perpendicular bisector."
          ],
          "hint": "The perpendicular bisector is the locus of points equidistant from the two endpoints. What does the compass have to do to guarantee equal distances?",
          "diagram": "<svg viewBox=\"0 0 240 164\" width=\"100%\" style=\"max-width:240px; height:auto; display:block; margin:0 auto;\" role=\"img\"><line x1=\"50\" y1=\"80\" x2=\"190\" y2=\"80\" stroke=\"#7c3aed\" stroke-width=\"2.2\" stroke-linecap=\"round\"/><circle cx=\"50\" cy=\"80\" r=\"3.2\" fill=\"#7c3aed\"/><circle cx=\"190\" cy=\"80\" r=\"3.2\" fill=\"#7c3aed\"/><circle cx=\"50\" cy=\"80\" r=\"90\" fill=\"none\" stroke=\"#6b6b76\" stroke-width=\"1.2\"/><circle cx=\"190\" cy=\"80\" r=\"90\" fill=\"none\" stroke=\"#6b6b76\" stroke-width=\"1.2\"/><line x1=\"120\" y1=\"4\" x2=\"120\" y2=\"156\" stroke=\"#c0392b\" stroke-width=\"1.5\" stroke-dasharray=\"5 4\" stroke-linecap=\"round\"/><text x=\"40\" y=\"96\" fill=\"#6b6b76\" font-size=\"11\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">A</text><text x=\"192\" y=\"96\" fill=\"#6b6b76\" font-size=\"11\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">B</text><text x=\"126\" y=\"20\" fill=\"#c0392b\" font-size=\"10.5\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">⟂ bisector</text></svg>",
          "diagramCaption": "Equal radii from both endpoints. Not drawn to scale"
        },
        {
          "id": "ccp-009",
          "topic": "Constructions",
          "difficulty": "medium",
          "stem": "To inscribe a circle inside a triangle, which construction locates the centre?",
          "choices": [
            "The intersection of the three angle bisectors",
            "The intersection of the three perpendicular bisectors",
            "The intersection of the three medians",
            "The intersection of the three altitudes"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. The angle bisectors meet at the incentre, which is equidistant from all three SIDES — exactly what a circle tangent to all three sides needs. The radius is the perpendicular distance from that point to any side.",
            "Perpendicular bisectors meet at the circumcentre, which is equidistant from the three VERTICES. That gives the circle THROUGH the corners, not the one inside touching the edges.",
            "Medians meet at the centroid, the balance point. It has no equidistance property.",
            "Altitudes meet at the orthocentre, which has no equidistance property either and can fall outside the triangle."
          ],
          "hint": "An inscribed circle touches the sides. Which centre is equidistant from the sides rather than the corners?"
        },
        {
          "id": "ccp-010",
          "topic": "Constructions",
          "difficulty": "hard",
          "stem": "You are given a circle and a point P outside it, and must construct a tangent from P. Why is the circle with diameter OP (where O is the centre) the key step?",
          "choices": [
            "Because any angle inscribed in a semicircle is a right angle, so where it meets the original circle the radius is perpendicular to PT",
            "Because it bisects the segment OP",
            "Because it has the same radius as the original circle",
            "Because it passes through the centre O"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. A tangent is perpendicular to the radius at the point of tangency, so the tangent point T must satisfy ∠OTP = 90°. Points seeing OP at a right angle lie exactly on the circle with OP as diameter — Thales' theorem — so intersecting that circle with the original one finds T.",
            "It does bisect OP, in the sense that its centre is OP's midpoint, but that is a by-product of the construction rather than the reason it works.",
            "It generally does not have the same radius, and the radius is not what makes the construction work.",
            "It passes through O, but so do infinitely many circles. What matters is the right angle the DIAMETER guarantees."
          ],
          "hint": "What angle must the radius make with the tangent at the point of contact? Then ask which locus of points sees a fixed segment at that angle."
        },
        {
          "id": "ccp-011",
          "topic": "Circles on the Coordinate Plane",
          "difficulty": "easy",
          "stem": "What is the measure of an angle inscribed in a semicircle?",
          "choices": [
            "90°",
            "60°",
            "180°",
            "It depends on the radius"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. A semicircle is a 180° arc, and an inscribed angle is half its intercepted arc, so the angle is 90°. This is Thales' theorem, and it is what makes the tangent construction from an external point work.",
            "60° would come from a 120° arc, not a semicircle.",
            "180° is the ARC, not the inscribed angle standing on it. The angle is half the arc.",
            "It does not depend on the radius. Every angle inscribed in a semicircle is a right angle, whatever the circle's size."
          ],
          "hint": "Halve the arc. A semicircle is how many degrees?"
        },
        {
          "id": "ccp-012",
          "topic": "Coordinate Geometry History",
          "difficulty": "easy",
          "stem": "Who published 'La Géométrie' in 1637, introducing the Cartesian coordinate system that bridges algebra and geometry?",
          "choices": [
            "René Descartes",
            "Johann Lambert",
            "Euclid",
            "Leonhard Euler"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. Descartes' La Géométrie fused algebra with geometry by putting figures on a coordinate grid — which is why the plane bears his name and why a circle can be written as an equation at all.",
            "Johann Lambert worked in the 18th century on non-Euclidean geometry and on proving π irrational.",
            "Euclid wrote the Elements around 300 BC, roughly two thousand years earlier and entirely without coordinates.",
            "Euler was an 18th-century mathematician — the Euler Line is his. La Géométrie predates him by a century."
          ],
          "hint": "The adjective for the coordinate plane is taken from this person's name."
        },
        {
          "id": "ccp-013",
          "topic": "Equation of a Circle",
          "difficulty": "easy",
          "stem": "Give the center and radius of the circle (x − 3)² + (y + 2)² = 25.",
          "choices": [
            "Center (3, −2), radius 5",
            "Center (−3, 2), radius 5",
            "Center (3, −2), radius 25",
            "Center (3, 2), radius 5"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. In the form (x − h)² + (y − k)² = r², the center is (h, k) and the radius is √r². Here (y + 2) reads as (y − (−2)), so k = −2, and r = √25 = 5.",
            "This flips both signs. The signs in the equation are already reversed relative to the center coordinates, so reversing them again returns you to the wrong place.",
            "The center is right but 25 is r², not r. The square root step was skipped.",
            "The x-coordinate is right but the y-coordinate keeps the sign as written. Since the form subtracts k, a plus sign in the equation means a negative coordinate."
          ],
          "hint": "The standard form subtracts the center coordinates, so read the signs as reversed — and remember the right side is r squared."
        },
        {
          "id": "ccp-014",
          "topic": "Completing the Square",
          "difficulty": "hard",
          "stem": "Find the radius of the circle x² + y² − 6x + 8y + 9 = 0.",
          "choices": [
            "3",
            "4",
            "5",
            "9"
          ],
          "correctIndex": 1,
          "explanations": [
            "3 comes from taking √9 using the constant term as written. That constant has to be moved and adjusted by the completing-the-square step before it means anything.",
            "Correct. Group and complete the square: (x² − 6x) + (y² + 8y) = −9 becomes (x − 3)² − 9 + (y + 4)² − 16 = −9, so (x − 3)² + (y + 4)² = 16 and r = 4.",
            "5 would follow from r² = 25, which comes from adding 9 and 16 without accounting for the −9 already on the right side.",
            "9 is the original constant term, not the radius. It changes value during the completing-the-square process."
          ],
          "hint": "Complete the square on x and y separately, tracking carefully what each step adds to both sides."
        },
        {
          "id": "ccp-015",
          "topic": "Parabola Definition",
          "difficulty": "medium",
          "stem": "A parabola is defined as the set of all points equidistant from which two objects?",
          "choices": [
            "Two fixed points (the foci)",
            "A fixed point and a fixed line",
            "Two fixed lines",
            "A fixed point and a circle"
          ],
          "correctIndex": 1,
          "explanations": [
            "Equidistance from two fixed points gives the perpendicular bisector of the segment joining them — a straight line. A constant SUM of distances to two foci gives an ellipse, and a constant difference gives a hyperbola.",
            "Correct. A parabola is the locus of points equidistant from a fixed point (the focus) and a fixed line (the directrix). Every point on the curve has the property that its distance to the focus equals its perpendicular distance to the directrix.",
            "Points equidistant from two lines lie on an angle bisector, or on a line midway between them if the two are parallel. Neither is a parabola.",
            "Equidistance from a point and a circle produces an ellipse or hyperbola depending on the configuration, not a parabola."
          ],
          "hint": "One of the two defining objects is zero-dimensional and the other is one-dimensional. Name each."
        },
        {
          "id": "ccp-016",
          "topic": "Circle Through a Point",
          "difficulty": "medium",
          "stem": "A circle has center (2, −1) and passes through the point (5, 3). What is its equation?",
          "choices": [
            "(x − 2)² + (y + 1)² = 5",
            "(x − 2)² + (y + 1)² = 25",
            "(x + 2)² + (y − 1)² = 25",
            "(x − 5)² + (y − 3)² = 25"
          ],
          "correctIndex": 1,
          "explanations": [
            "The center is placed correctly but the right side holds r rather than r². The radius here is 5, so the constant must be 25.",
            "Correct. The radius is the distance from the center to the given point: √((5−2)² + (3−(−1))²) = √(9 + 16) = √25 = 5. With center (2, −1), the equation is (x − 2)² + (y + 1)² = 5² = 25.",
            "This reverses the signs of the center coordinates. Standard form subtracts the center, so a center of (2, −1) produces (x − 2) and (y + 1).",
            "This centers the circle on the point it passes through rather than on the given center. The two roles are not interchangeable."
          ],
          "hint": "Find the radius with the distance formula first, then place the center into standard form — watching both the signs and the squaring."
        },
        {
          "id": "ccp-017",
          "topic": "Perpendicular Bisector Construction",
          "difficulty": "hard",
          "stem": "In the classical compass-and-straightedge construction of a perpendicular bisector, arcs of equal radius are drawn from both endpoints of a segment. Why does the line through the two arc intersections bisect the segment perpendicularly?",
          "choices": [
            "Because the arcs are drawn with a radius exactly half the segment.",
            "Because each intersection point is equidistant from both endpoints, and that locus is the perpendicular bisector.",
            "Because the two arcs are congruent circles.",
            "Because the compass width is arbitrary."
          ],
          "correctIndex": 1,
          "explanations": [
            "The radius need not be half the segment — in fact it must be MORE than half or the arcs never meet. Any radius above that threshold works equally well, which is a hint that the specific value is not what makes the construction succeed.",
            "Correct. Each intersection point sits at the same distance from both endpoints, because both arcs were struck with the same compass setting. The set of all points equidistant from two fixed points IS the perpendicular bisector, so two such points determine it.",
            "The arcs being congruent is what makes each intersection equidistant from the endpoints, so this is a necessary ingredient — but it is not the reason on its own. The conclusion follows from the equidistance locus, not from congruence as such.",
            "The compass width is not entirely arbitrary; it must exceed half the segment length for the arcs to intersect at all. And even granting a valid width, arbitrariness explains nothing about why the result is perpendicular."
          ],
          "hint": "Ask what property the two intersection points share, then recall which locus is defined by exactly that property."
        },
        {
          "id": "ccp-018",
          "topic": "Reasoning: Identify the Conic",
          "difficulty": "hard",
          "stem": "An equation has both x² and y² terms with the SAME coefficient and opposite... no, the same sign, and no xy term. What conic is it, and what would change that?",
          "choices": [
            "A parabola; adding a y² term would make it a circle.",
            "A circle. If the coefficients were unequal but still the same sign it would be an ellipse, and with opposite signs a hyperbola.",
            "A hyperbola, because two squared terms always separate the branches.",
            "It cannot be determined without the constant term."
          ],
          "correctIndex": 1,
          "explanations": [
            "A parabola has only ONE squared variable. Two squared terms rule it out immediately.",
            "Correct. Equal coefficients with the same sign give a circle; unequal but same sign gives an ellipse; opposite signs give a hyperbola; only one squared term gives a parabola.",
            "Opposite signs produce a hyperbola. The same sign keeps the curve closed.",
            "The constant affects the size, and can make the graph empty or a single point, but the TYPE is decided by the squared terms."
          ],
          "hint": "Compare the two squared coefficients: same or different, and same sign or opposite?",
          "kind": "reasoning"
        },
        {
          "id": "ccp-019",
          "topic": "Reasoning: Error Analysis",
          "difficulty": "medium",
          "stem": "A student writes that x² + y² = −16 is a circle of radius −4. What is actually true?",
          "choices": [
            "It is a circle of radius 4, since radius is always taken positive.",
            "No real graph exists — a sum of two squares cannot be negative.",
            "It is a hyperbola.",
            "It is a single point at the origin."
          ],
          "correctIndex": 1,
          "explanations": [
            "Flipping the sign changes the equation into a different one. As written, nothing satisfies it.",
            "Correct. x² and y² are each greater than or equal to zero, so their sum can never be −16. The equation has no real solutions and therefore no graph.",
            "A hyperbola needs opposite signs on the squared terms. Both are positive here.",
            "x² + y² = 0 is the single point at the origin. A negative right-hand side gives nothing at all."
          ],
          "kind": "reasoning"
        }
      ]
    }
  }
}