If a secant segment and a tangent segment share an endpoint outside a circle, then the product of the lengths of the secant segment and its external segment equals the square of the length of the tangent segment.
First, we need to calculate the lengths of the secant SP by adding x and 10.
SP=x+10
According to this theorem, the product of SP and TP is equal to the square of PR.
SP* TP =PR^2
Let's substitute SP with x+10, TP with x, and PR with 12 and solve the equation for x.
This is a quadratic equation, so let's solve it by completing the square. Note that all terms with x are on one side of the equation, so we do not need to rewrite it. In this equation, b=10. Using this information, we can calculate ( b2 )^2.
We conclude that the equation has two solutions.
x_1&=- 5+13=8
x_2&=- 5-13=- 18
On the diagram, x represents the distance TP, so it cannot have a negative value. Therefore, x is equal to 8.
