Bluebook SAT Test Circles 5

Question 1

The equation 2x² + 2y² − 12x + 16y + 10 = 0 defines a circle. What is the area of this circle?

Accepted Answer

20π

Answer

First divide every term by 2: x² + y² − 6x + 8y + 5 = 0. Complete the square for each variable: (x² − 6x + 9) + (y² + 8y + 16) = −5 + 9 + 16, giving (x − 3)² + (y + 4)² = 20. The radius squared is 20, so the area is π · 20 = 20π. A common trap is keeping the leading coefficient of 2, which would incorrectly double r² to 40.

Question 2

From external point P(12, 5), a tangent segment is drawn to a circle centered at the origin. The tangent segment has length 12. What is the radius of the circle?

Accepted Answer

5

Answer

The distance from P(12, 5) to the origin is √(12² + 5²) = √(144 + 25) = √169 = 13. For a tangent from an external point, the relationship is: tangent² + radius² = distance². So 12² + r² = 13², giving r² = 169 − 144 = 25, and r = 5. The 5–12–13 Pythagorean triple is deliberately obscured here.

Question 3

Two chords AB and CD intersect at point P inside a circle. AP = 3, PB = 12, and PD is 5 more than CP. What is the total length of chord CD?

Accepted Answer

13

Answer

By the intersecting chords theorem, AP · PB = CP · PD. Let CP = x, so PD = x + 5. Then 3 · 12 = x(x + 5), giving x² + 5x − 36 = 0. Factoring: (x + 9)(x − 4) = 0, so x = 4 (taking the positive value). Thus CP = 4 and PD = 9, making CD = 4 + 9 = 13.

Question 4

In a circle, chord AB is parallel to chord CD. If arc AC = 50°, what is the measure of arc BD?

Accepted Answer

50°

Answer

When two chords are parallel, the arcs between them on each side are equal: arc AC = arc BD. This is because the parallel chords create congruent alternate interior angles with any transversal, which correspond to equal inscribed angles — and equal inscribed angles intercept equal arcs. Therefore arc BD = 50°.

Question 5

From external point P, a tangent PA touches a circle at A, and a secant from P passes through the circle at points B (nearer) and C (farther). If PA = 8 and PB = 4, what is the length of chord BC?

Accepted Answer

12

Answer

By the power of a point theorem for a tangent and secant from the same external point: PA² = PB · PC. So 64 = 4 · PC, giving PC = 16. The chord BC = PC − PB = 16 − 4 = 12. A common error is stopping at PC = 16, which is the full secant length from P to C, not the chord length.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Circles 5