Point W lies outside circle V. A secant from W passes through points A and B on the circle (A closer to W), with WA = 8 and AB = 10. Another line from W is tangent to the circle at point C. If a third secant from W passes through points D and E (D closer to W) with WD = 6, what is the length DE?
Answer: 18
Use the secant-secant theorem with the two secants: WA × WB = WD × WE. We have WB = WA + AB = 8 + 10 = 18. So 8 × 18 = 6 × WE, giving 144 = 6 × WE, thus WE = 24. Therefore DE = WE - WD = 24 - 6 = 18. Choice B might result from adding WA + AB incorrectly. Choice C could come from miscalculation in the proportion. Choice D incorrectly uses WE as the answer.
In circle O, two chords AB and CD intersect at point P inside the circle. If AP = 6, PB = 4, and CP = 8, what is the length of PD?
Answer: 3
When two chords intersect inside a circle, the products of their segments are equal: AP × PB = CP × PD. Substituting: 6 × 4 = 8 × PD, so 24 = 8 × PD, therefore PD = 3. Choice B uses PB incorrectly. Choice C assumes the segments form a 3-4-5 pattern. Choice D incorrectly uses AP as the answer.
Two secants are drawn from external point W to circle O. One secant intersects the circle at X and Y (X between W and Y), and the other intersects at points Z and V (Z between W and V). If WX = 6, XY = 10, and WZ = 8, what is WV?
Answer: 12
For two secants from an external point: WX × WY = WZ × WV. First, WY = WX + XY = 6 + 10 = 16. Then 6 × 16 = 8 × WV, so 96 = 8 × WV, giving WV = 12. Choice B incorrectly uses WV = WX + XY - 1. Choice C uses WV = WY + 2. Choice D assumes WV = WX + XY + WZ - 2.
In circle O, two chords AB and CD intersect at point P inside the circle. If AP = 6, PB = 8, and CP = 4, what is the length of PD?
Answer: 12
When you see two chords intersecting inside a circle, you're dealing with the intersecting chords theorem. This powerful geometric principle states that when two chords intersect at a point inside a circle, the products of their segments are equal.
For chords AB and CD intersecting at point P, the theorem tells us that AP×PB=CP×PD. This relationship holds because the angles formed create similar triangles, making the cross-products equal.
Let's apply this to our problem. We know AP = 6, PB = 8, and CP = 4, and we need to find PD. Setting up the equation:
6×8=4×PD48=4×PDPD=12
Looking at the wrong answers: Choice A (14) might result from incorrectly adding AP + PB rather than multiplying them. Choice B (10) could come from mistakenly calculating CPAP×PB=448−2=10 through some algebraic error. Choice D (16) might arise from incorrectly thinking the relationship is AP+PB=CP+PD, which would give 6+8=4+PD, so PD=10, but then doubling due to confusion.
Study tip: Memorize the intersecting chords theorem as "segment times segment equals segment times segment." Whenever you see chords crossing inside a circle with three given lengths, immediately set up the cross-multiplication equation. This pattern appears frequently in geometry problems.
A tangent from point T touches circle M at point N, and a secant from T passes through points P and Q on the circle (with P closer to T). If TN = 12 and TP = 8, what is the length of PQ?
Answer: 10
For a tangent and secant from the same external point, TN² = TP × TQ. So 12² = 8 × TQ, giving 144 = 8 × TQ, thus TQ = 18. Therefore PQ = TQ - TP = 18 - 8 = 10. Choice A incorrectly uses TP. Choice C uses TN as the answer. Choice D incorrectly uses TQ.
From point A outside circle O, a tangent and a secant are drawn. The tangent touches the circle at B, and the secant passes through points C and D on the circle (C is between A and D). If AB = 221 and AC = 7, what is the length of AD?
Answer: 12
Using the tangent-secant theorem: AB² = AC × AD. We have (2√21)² = 7 × AD, so 4 × 21 = 7 × AD, giving 84 = 7 × AD, therefore AD = 12. Choice B incorrectly adds AB and AC. Choice C uses an incorrect doubling of AB. Choice D uses AB² ÷ 4 instead of the correct relationship.
A tangent from external point T touches circle O at point A, and a secant from T passes through points B and C on the circle (with B between T and C). If TA = 12 and TB = 8, what is the length of TC?
Answer: 18
For a tangent and secant from the same external point, TA² = TB × TC. So 12² = 8 × TC, which gives 144 = 8 × TC, therefore TC = 18. Choice A incorrectly uses TA = TB + BC. Choice C assumes TC = TB + TA incorrectly. Choice D uses 2 × TA instead of TA².
From external point T, two tangent segments are drawn to circle O, touching at points U and V. A secant from T intersects the circle at points W and X (W between T and X). If TU = 20 and TW = 16, what is TX?
Answer: 25
Since tangent segments from an external point are equal, TV = TU = 20. Using the tangent-secant theorem: TU² = TW × TX. So 20² = 16 × TX, giving 400 = 16 × TX, therefore TX = 25. Choice A incorrectly uses TX = TU + 4. Choice C uses TX = TW + TU - 6. Choice D assumes TX = TU + TW incorrectly.
From an external point Q, two secants are drawn to circle R. One secant passes through points A and B (with A closer to Q), and the other passes through points C and D (with C closer to Q). If QA = 5, AB = 7, and QC = 4, what is the length of CD?
Answer: 9
For two secants from an external point, QA × QB = QC × QD. First, QB = QA + AB = 5 + 7 = 12. Then 5 × 12 = 4 × QD, so QD = 15. Therefore CD = QD - QC = 15 - 4 = 9. Choice A uses AB incorrectly. Choice C adds QA and AB incorrectly. Choice D uses QB as the answer.
Two chords of circle O intersect at point K inside the circle. One chord has segments of length a and b on either side of K, while the other has segments of length c and d. If a:b = 2:3 and c:d = 4:1, and the product ab = 24, what is the value of cd?
Answer: 24
Since the chords intersect inside the circle, ab = cd. Given ab = 24, we have cd = 24. The ratios a:b = 2:3 and c:d = 4:1 are additional information that confirms our segments are valid, but the key theorem is that the products must be equal. Choice A might come from incorrectly using the ratio 4:1 to get 4×5=20. Choice C could result from adding the ratio parts. Choice D might come from using 2×16=32.
From point M outside circle N, a tangent MT is drawn touching the circle at T, and a secant is drawn through points X and Y on the circle (X closer to M). If MT = 15 and the secant has total length MY = 25, what is the length MX?
Answer: 9
Using the tangent-secant theorem: MT² = MX × MY. So 15² = MX × 25, giving 225 = 25 × MX, therefore MX = 9. Choice B would result from a calculation error. Choice C incorrectly uses the tangent length. Choice D comes from incorrectly using 16 × 25 = 400.
