Question

Given: Circle k(O), O∈

PL

KE
- tangent at E
KE=18, PL=15
Find: KP

Answer 1

The length of KP = 12 unit.

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.

`KE^2=KP\cdot KL`

KE = 18

KP = x

KL= x+15

`18^2=x(x+15)`

`x^2+15x-324=0`

`(x+27)(x-12)=0`

Set each factor to 0 and solve for x

`x+27=0\Rightarrow x=-27`

`x-12=0\Rightarrow x=12`

We will ignore negative value of x because length can't be negative.

Hence, The length of KP = 12 unit.

Answer 2

KP=12

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.