+876 Let E, F, G, and H be points on a circle such that EF = 15 and GH = 21. Point U is on segment \overline{EF} with EU = 8, and V is on segment \overline{GH} with GV = 5. Line UV intersects the circle at points X and Y. If UV = 6, then find XY.
I've got some very nasty equations plz help. :) thx!
+23252 Using this property: if two chords of a circle intersect, the product of the sections of one of the chords equals
the product of the sections of the other.
Since EF = 15 and EU = 8, UF = 7.
Since GH = 21 and GV = 5, HV = 16.
On chord XUVY, call section XU = x and VY = y.
Using chord EF and chord XUVY: 7·8 = x(y + 6).
Using chord GH an chord XUVY: 5·16 = y(x + 6)
Using 5·16 = y(x + 6) ---> 80 = y(x + 6) ---> y = 80 / (x + 6)
Using 7·8 = x(y + 6) ---> 56 = xy + 6x ---> xy = 56 - 6x ---> y = (56 - 6x) / x
Setting these two equations equal to each other: 80 / (x + 6) = (56 - 6x) / x
Cross-multiplying: (56 - 6x)(x + 6) = 80x
Simplifying: x2 + 10x - 56 = 0
(x + 14)(x - 4) = 0
x = 4
From here, you can calculate both the value of y and the full length of the chord.
+129852 See the folllowing [ not to scale] image :
GV = 5
So VH = 21 - 5 = 16
And
EU = 8
So UF = 15 - 8 = 7
Call segment UX = m and segment VY = n
VX = 6 + m
UY = 6 + n
Using the intersecting chord theorem.....we have that
GV * VH = VX * VY
5 * 16 = (6 + m) n (1)
80 = 6n + mn
80 - 6n = mn
And
EU * UF = UY * UX
8 * 7 = (6 + n) m
56 = 6m + mn
56 - 6m = mn
This implies that
80 - 6n = 56 - 6m
80 - 56 = 6n - 6m
24 = 6(n - m)
4 = n - m
m + 4 = n (2)
So using (1) and (2) we have that
80 = (6 + m) (m + 4)
80 = m^2 + 10m + 24
m^2 + 10m - 56 = 0
(m + 14) ( m - 4) = 0
Setting both factors to 0 and solving for m produces m = -14 (reject) or m = 4
So
m + 4 = n
4 + 4 = n
8 = n
So.....the length of XY is
m + UV + n =
4 + 6 + 8 =
18
