Question

in the accompanying diagram, PA is tangent to the circle at A and PBC is a seacant. If PA= 2 square root 3 and PB =2, what is PC?

Answer 1

PC = 6

Step-by-step explanation:

PA = 2 square root 3 = 2√3

PB = 2

PC = ?

We would apply secant-tangent theorem:

`PA^2\text{ = }PB\text{ }*\text{ PC}`
`\begin{gathered} (2\surd3)^2\text{ = 2(PC)} \\ (2\surd3)(2\surd3)\text{ = 2PC} \\ 4(\sqrt[]{3})^2\text{ = 2PC} \\ 4(3)\text{ = 2PC} \end{gathered}`
`\begin{gathered} 12\text{ = 2PC} \\ \text{divide both sides by 2:} \\ (12)/(2)=(2PC)/(2) \\ PC\text{ = 6} \end{gathered}`