Grade 10 Circles & Analytic Geometry Practice — Hard

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Question 1 of 17: In a cyclic quadrilateral (inscribed in a circle), opposite angles are:

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Answer key for parents & teachers (17 questions)
  1. In a cyclic quadrilateral (inscribed in a circle), opposite angles are:Supplementary (sum to 180°). In a cyclic quadrilateral, opposite angles are supplementary — they sum to 180°. This is a key theorem.
  2. A chord of a circle is 90° from the center. The inscribed angle intercepting the same chord as a diameter is:90°. Any inscribed angle in a semicircle (intercepting a diameter) = 90°. This is Thales' theorem.
  3. A circle has radius 5 cm and sector area 10π cm². What is the central angle in radians?2π/5. A = (1/2)r²θ → 10π = (1/2)(25)θ → θ = 20π/25 = 4π/5. Wait: (1/2)(25)(θ) = 10π → 12.5θ = 10π → θ = 10π/12.5 = 4π/5. So angle = 4π/5 radians.
  4. A sector has area 36π and radius 6. What is the central angle?2π radians. A = (1/2)r²θ → 36π = (1/2)(36)θ = 18θ → θ = 2π radians (= 360°). This is the full circle!
  5. A circle has equation (x − 4)² + (y + 7)² = 81. Does the point (4, 2) lie inside, on, or outside?Inside (distance < r). Distance from center (4, −7) to (4, 2): √((4−4)² + (2−(−7))²) = √81 = 9 = r. Point is ON the circle!
  6. The equation (x − 1)² + (y − 2)² = −5 describes:No real circle exists. r² must be positive. r² = −5 < 0 is impossible. This equation has no real graph.
  7. If r² = 0 in the standard equation, the graph is:A point (degenerate circle). r² = 0 means r = 0 — a circle of zero radius, which is just a single point (the center). This is a degenerate circle.
  8. Find the radius of x² + y² − 6x − 8y = 0.r = 5. Complete the square: (x−3)² + (y−4)² = 9+16 = 25. r = 5.
  9. A circle passes through (0, 0) and has center (3, 4). What is the equation?(x−3)² + (y−4)² = 25. r = distance from center (3,4) to (0,0) = √(9+16) = 5. r² = 25. Equation: (x−3)² + (y−4)² = 25.
  10. x² + y² + 10x − 4y + 20 = 0. Find the center.(−5, 2). Group: (x²+10x) + (y²−4y) = −20. Complete: (x+5)² + (y−2)² = −20+25+4 = 9. Center: (−5, 2).
  11. The general form of a circle equation is x² + y² + Dx + Ey + F = 0. The center is:(−D/2, −E/2). Completing the square: center = (−D/2, −E/2). The negative signs come from the standard form (x−h)².
  12. A strategic choice for placing a triangle in a coordinate proof is:One vertex at the origin, one side along the x-axis. Placing one vertex at the origin and a side along the x-axis gives simpler coordinates (like (0,0), (a,0), (b,c)) making algebra cleaner.
  13. In a general coordinate proof, vertices use variables like (a, 0) and (0, b) to:Keep the proof general for ALL such triangles. General coordinates make the proof valid for ALL triangles of that type, not just one specific example.
  14. The midpoint of the hypotenuse of a right triangle with vertices (0,0), (2a,0), (0,2b) is:(a, b). Hypotenuse goes from (2a, 0) to (0, 2b). Midpoint = ((2a+0)/2, (0+2b)/2) = (a, b).
  15. To prove a quadrilateral is a RHOMBUS using coordinates:Show all four sides have equal length. A rhombus has all four sides equal. Use the distance formula to show all four side lengths are equal.
  16. To prove a triangle is a RIGHT triangle using coordinates:Show two sides have perpendicular slopes (slopes are negative reciprocals). Prove two sides are perpendicular using slopes: m₁ × m₂ = −1. Those sides form the right angle.
  17. The centroid (average of vertices) of a triangle with vertices A(x₁,y₁), B(x₂,y₂), C(x₃,y₃) is:((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3). The centroid is the average of all three vertices: ((x₁+x₂+x₃)/3, (y₁+y₂+y₃)/3).