Question

(a) in the figure below, two secants are drawn to a circle from exterior point e. suppose that ew = 68, ey = 51, and ez = 25.5. find xy. xy = \boxed{}

Explanation:

Step1: Recall the secant-secant rule

The secant-secant rule states that if two secant segments are drawn from a point outside a circle, then \( EY \times EX = EZ \times EW \). Let \( XY = x \), so \( EX = EY - XY = 51 - x \) (wait, no, actually \( EX = EY - XY \) is wrong. Wait, \( EY \) is the length from \( E \) to \( Y \), and \( EX \) is from \( E \) to \( X \), so \( EY = EX + XY \), so \( EX = EY - XY \)? Wait, no, the diagram: \( E \) is outside, \( X \) and \( Y \) are on the circle, with \( EY \) passing through \( X \)? Wait, no, the two secants: one is \( EY \) (from \( E \) to \( Y \), with \( X \) between \( E \) and \( Y \)) and the other is \( EW \) (from \( E \) to \( W \), with \( Z \) between \( E \) and \( W \)). So the correct formula is \( EY \times EX = EZ \times EW \)? Wait, no, the secant-secant theorem is: if a secant from \( E \) passes through the circle at \( X \) and \( Y \) (so \( EX \) is the external segment, \( XY \) is the internal segment, so total \( EY = EX + XY \)), and another secant passes through \( Z \) and \( W \), so \( EZ \) is external, \( ZW \) is internal, total \( EW = EZ + ZW \). Then the theorem is \( EX \times EY = EZ \times EW \). Wait, no, the correct formula is \( (external)(external + internal) = (external)(external + internal) \). So \( EX \times EY = EZ \times EW \), where \( EX \) is the length from \( E \) to the first intersection \( X \), \( EY \) is from \( E \) to the second intersection \( Y \), so \( EY = EX + XY \), and \( EW = EZ + ZW \). So given \( EW = 68 \), \( EY = 51 \), \( EZ = 25.5 \). Let \( EX = a \), \( XY = b \), so \( EY = a + b = 51 \), and \( EW = EZ + ZW = 25.5 + ZW = 68 \), so \( ZW = 68 - 25.5 = 42.5 \). Then by the secant-secant theorem: \( EX \times EY = EZ \times EW \), so \( a \times 51 = 25.5 \times 68 \). Wait, no, that's not right. Wait, the correct formula is \( (length of external part) \times (length of entire secant) = (length of external part of other secant) \times (length of entire other secant) \). So for secant \( EYW \) (wait, no, the two secants are \( EXY \) and \( EZW \), where \( E \) is outside, \( X \) and \( Y \) are on the circle (so \( EX \) is external, \( XY \) is internal, \( EY = EX + XY \)), and \( Z \) and \( W \) are on the circle ( \( EZ \) is external, \( ZW \) is internal, \( EW = EZ + ZW \)). Then the theorem is \( EX \times EY = EZ \times EW \). Wait, let's check the values. Let \( EX = x \), so \( EY = x + XY \), but we know \( EY = 51 \), so \( x + XY = 51 \), so \( XY = 51 - x \). For the other secant, \( EZ = 25.5 \), \( EW = 68 \), so \( EZ \times EW = 25.5 \times 68 \). And \( EX \times EY = x \times 51 \). So set them equal: \( x \times 51 = 25.5 \times 68 \). Solve for \( x \): \( x = \frac{25.5 \times 68}{51} \). Calculate that: \( 25.5 \times 68 = 25.5 \times 68 \). Let's compute \( 25.5 \times 68 \): \( 25 \times 68 = 1700 \), \( 0.5 \times 68 = 34 \), so total \( 1700 + 34 = 1734 \). Then \( x = \frac{1734}{51} = 34 \). So \( EX = 34 \), then \( XY = EY - EX = 51 - 34 = 17 \). Wait, that makes sense. Let's verify: \( EX \times EY = 34 \times 51 = 1734 \), \( EZ \times EW = 25.5 \times 68 = 1734 \). Yes, that works. So \( XY = 17 \).

Step2: Apply the secant-secant formula

The secant-secant theorem formula is \( EZ \times EW = EX \times EY \). Let \( EX = y \), then \( y \times 51 = 25.5 \times 68 \). Solving for \( y \): \( y = \frac{25.5 \times 68}{51} \). Calculate \( 25.5 \div 51 = 0.5 \), so \( 0.5 \times 68 = 34 \). So \( EX = 34 \). Then \( XY = EY - EX = 51 - 34 = 17 \).

Answer:

\( \boxed{17} \)