Circles Flashcards
6 cards from real Bluebook SAT Test practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 6 Circles flashcards as text
Which of the following equations does NOT represent a circle in the coordinate plane? A) x² + y² − 6x + 8y + 9 = 0 B) x² + y² + 4x − 2y + 5 = 0 C) x² + y² − 2x + 6y − 6 = 0 D) x² + y² + 10x − 8y + 5 = 0
Answer: x² + y² + 4x − 2y + 5 = 0
Complete the square for each equation. For option B: (x² + 4x + 4) + (y² − 2y + 1) + 5 − 4 − 1 = 0, which gives (x + 2)² + (y − 1)² = 0. The only solution is the single point (−2, 1). Since the 'radius squared' equals 0, this is a degenerate case — a single point, not a circle. The other equations all yield positive radii: A gives radius 4, C gives radius 4, and D gives radius 6.
Two chords AB and CD intersect at point P inside a circle. If AP = 3, CP = 4, and PD = 6, what is the length of chord AB?
Answer: 11
By the Intersecting Chords Theorem, AP × PB = CP × PD. Substituting: 3 × PB = 4 × 6 = 24, so PB = 8. The full length of chord AB = AP + PB = 3 + 8 = 11. A common error is choosing 8, which is only the length of segment PB, not the entire chord.
From external point P, two tangent segments are drawn to a circle, touching the circle at points A and B. If angle APB = 60° and the radius of the circle is 6, what is the length of each tangent segment PA?
Answer: 6√3
Let O be the center. Since PA and PB are tangents, OA ⊥ PA and the figure is symmetric, so ∠APO = 30°. In right triangle OAP, OA = 6 (radius) and ∠OAP = 90°. Using tan(30°) = OA/PA → (1/√3) = 6/PA → PA = 6√3. Alternatively, sin(30°) = 6/OP gives OP = 12, then PA = √(OP² − OA²) = √(144 − 36) = √108 = 6√3.
Points A, B, C, and D lie on a circle in that order. Arc AB = 70°, arc BC = 90°, and arc CD = 80°. What is the measure of inscribed angle BAD?
Answer: 85°
First find arc DA: 360° − 70° − 90° − 80° = 120°. Inscribed angle BAD has its vertex at A and intercepts the arc from B to D that does NOT contain A, which is arc BCD = arc BC + arc CD = 90° + 80° = 170°. By the Inscribed Angle Theorem, ∠BAD = 170°/2 = 85°. A typical error is using arc AB + arc DA = 70° + 120° = 190°, which intercepts the wrong arc.
A sector of a circle has a perimeter of 36 units. If the radius of the circle is 10 units, what is the area of the sector?
Answer: 80
The perimeter of a sector consists of two radii plus the arc length: Perimeter = 2r + arc length. Substituting: 36 = 2(10) + arc length → arc length = 16. The area of a sector can be expressed as A = (1/2) × r × arc length = (1/2)(10)(16) = 80 square units. This elegant formula A = ½rl (where l is arc length) avoids needing the central angle in degrees.
For what values of m is the line y = mx + 4 tangent to the circle x² + y² = 4?
Answer: m = ±√3
Substitute y = mx + 4 into x² + y² = 4: x² + (mx + 4)² = 4 → x²(1 + m²) + 8mx + 12 = 0. For the line to be tangent (exactly one intersection point), the discriminant must equal zero: Δ = (8m)² − 4(1 + m²)(12) = 0 → 64m² − 48 − 48m² = 0 → 16m² = 48 → m² = 3 → m = ±√3. Note the line y = 4 (m = 0) does not intersect the circle of radius 2 centered at the origin since it lies above it.