Practice Test GeeksBluebook SAT Practice Test
← All Bluebook SAT Test Flashcard Decks

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

  1. The equation x² + y² + 6x − 4y + k = 0 represents a circle with area 12π. What is the value of k?

    Answer: 1

    Complete the square: (x² + 6x + 9) + (y² − 4y + 4) = −k + 9 + 4, giving (x + 3)² + (y − 2)² = 13 − k. For area = 12π, we need πr² = 12π, so r² = 12. Setting 13 − k = 12 gives k = 1.

  2. Point P lies outside a circle. A tangent from P touches the circle at T with PT = 6. A secant from P enters the circle at A and exits at B, with A between P and B. If PA = 4, what is the length of chord AB?

    Answer: 5

    By the Power of a Point theorem, PT² = PA · PB. So 36 = 4 · PB, giving PB = 9. Since A is between P and B, AB = PB − PA = 9 − 4 = 5.

  3. A sector of a circle has a perimeter of 30 and a radius of 8. What is the area of the sector?

    Answer: 56

    The perimeter of a sector is 2r + arc length. So 2(8) + arc = 30, giving arc = 14. The area of a sector equals ½ · r · arc = ½ · 8 · 14 = 56.

  4. Cyclic quadrilateral ABCD is inscribed in a circle. The arcs measure: arc AB = 60°, arc BC = 120°, arc CD = 90°, and arc DA = 90°. What is the measure of inscribed angle DAB?

    Answer: 105°

    Inscribed angle DAB subtends arc BCD (the arc from B to D not passing through A). Arc BCD = arc BC + arc CD = 120° + 90° = 210°. By the Inscribed Angle Theorem, angle DAB = 210° / 2 = 105°.

  5. For which values of m does the line y = mx + 5 intersect the circle x² + y² = 9 at exactly two distinct points?

    Answer: |m| > 4/3

    The distance from the center (0, 0) to the line mx − y + 5 = 0 is 5/√(m² + 1). For two intersections this distance must be less than the radius 3: 5/√(m² + 1) 16/9 → |m| > 4/3.

  6. Circle A is defined by x² + y² = 25 and Circle B is defined by (x − 7)² + y² = 4. What is the length of a common external tangent to these two circles?

    Answer: 2√10

    Circle A has center (0, 0) and radius 5; Circle B has center (7, 0) and radius 2. The distance between centers is d = 7. Since d = r₁ + r₂ = 5 + 2 = 7, the circles are externally tangent. The length of the external common tangent is √(d² − (r₁ − r₂)²) = √(49 − 9) = √40 = 2√10.