Question

10.7 special segments in circles.pdf

Explanation:

Step1: Apply the tangent - secant segment theorem

If a tangent and a secant are drawn from an external point to a circle, then \(20^{2}=x(x+(x - 6))\).

Step2: Expand the equation

$$
LATEXBLOCK0
$$

Step3: Rearrange to standard quadratic form

\(2x^{2}-6x - 400 = 0\), divide by 2: \(x^{2}-3x - 200 = 0\).

Step4: Solve the quadratic equation

Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), where \(a = 1\), \(b=-3\), \(c=-200\).

$$
LATEXBLOCK1
$$

We also can factor: \(x^{2}-3x - 200=(x - 16)(x + 13)=0\).

Answer:

\(x = 16\) (since \(x>0\) and \(x=-13\) is rejected as length cannot be negative)