Tangents Secants and Circles...Oh My

By the property of tangent and secant lines on a circle, we have

$\begin{aligned} \lvert\overline{PT}\rvert^2 &= \lvert\overline{PC}\rvert \times \lvert\overline{PD}\rvert \qquad (\text{ in circle } O')\\ \lvert\overline{PT}\rvert^2 &= \lvert\overline{PA}\rvert \times \lvert\overline{PB}\rvert. \qquad (\text{ in circle } O )\\ \end{aligned}$

Therefore, $\begin{aligned} \lvert\overline{PC}\rvert \times \lvert\overline{PD}\rvert &= \lvert\overline{PA}\rvert \times \lvert\overline{PB}\rvert\\ 30 \times (30 + 20) &= \lvert\overline{PA}\rvert \times \left(\lvert\overline{PA}\rvert + 35\right)\\ \lvert\overline{PA}\rvert^2 + 35 \times \lvert\overline{PA}\rvert &= 1500\\ \left(\lvert\overline{PA}\rvert - 25\right)\times \left(\lvert\overline{PA}\rvert + 60\right) &= 0. \\ \end{aligned}$

Since $\lvert\overline{PA}\rvert$ is positive, we have $\lvert\overline{PA}\rvert = 25.$

By the power of a point, we have

$(PT)^2=PA(PB)=PC(PD)$

$PA(PA+35)=30(50)$

$(PA)^2+35(PA)-1500=0$

Using the Quadratic Formula to compute for $PA$ , we have

$PA=\dfrac{-35\pm \sqrt{(-35)^2-4(-1500)}}{2}$

$PA=\dfrac{-35}{2}\pm \dfrac{85}{2}$

Considering the positive value only, we have

$PA=\dfrac{50}{2}=\boxed{25}$

Using properties of Secant Lines, -We know that:

(CD+CP) × CP = (BA + AP) × AP

-When We substitute the values we get:

(20 + 30) × 30 = (35 + AP) × AP

-Simplification:

1500= 35AP + AP^2

-This becomes a quadratic:

AP^2 + 35AP - 1500 =0

-The factors are -30 and 25, so we know that the length of AP(or PA) is 25

(AP+35) (AP-25)=0

By the power of a point on the lower circle, we have

$(PT)^2=(PC)(PD)=30(50)=1500$

By the power of a point on the upper circle, we have

$(PT)^2=(AP)(AP+35)$

$1500=(AP)^2+35(AP)$

$(AP)^2+35(AP)-1500=0$

$(AP+60)(AP-25)=0$

$\boxed{AP=25}$