Given: Circle k(O), O∈
PL
KE
- tangent at E
KE=18, PL=15
Find: KP

Answer:
KP=12
Step-by-step explanation:
Use property for circle and tangent segment:
tangent segment² = external secant segment · secant segment.
In your case,
tangent segment KE = 18;
external secant segment KP;
secant segment KL = KP + 15.
Thus,
18²=KP(KP+15),
KP²+15KP-324=0,
D=15^2-4·(-324)=225+1296=1521,
KP=(-15±39)/2=-27, 12.
The segment cannot be of negative length, then KP=12.
Answer:
The length of KP = 12 unit.
Step-by-step explanation:
Given: Circle with center O. KE is tangent at E and PL is diameter.
KE=18, PL=15 find KP
Let length of KP be x
Tangent- Secant Theorem: The square of length of tangent is equal to product of length of sub part of secant.
[tex]KE^2=KP\cdot KL[/tex]
KE = 18
KP = x
KL= x+15
[tex]18^2=x(x+15)[/tex]
[tex]x^2+15x-324=0[/tex]
[tex](x+27)(x-12)=0[/tex]
Set each factor to 0 and solve for x
[tex]x+27=0\Rightarrow x=-27[/tex]
[tex]x-12=0\Rightarrow x=12[/tex]
We will ignore negative value of x because length can't be negative.
Hence, The length of KP = 12 unit.