If two chords of a circle intersect inside the circle, the product of the lengths of the parts of each chord is the same. In other words,

\(\displaystyle 2 t \cdot t = 10 \cdot 10\)

\(\displaystyle 2 t ^{2}= 10 0\)

Solving for \(\displaystyle t\):

\(\displaystyle 2 t ^{2} \div 2 = 10 0 \div 2\)

\(\displaystyle t ^{2} = 50\)

\(\displaystyle t = \sqrt{50 }\)

Simplifying the radical using the Product of Radicals Principle, and noting that 25 is the greatest perfect square factor of 50:

\(\displaystyle t = \sqrt{25 } \cdot \sqrt{2} = 5 \sqrt{2}\)

If a secant segment and a tangent segment are constructed to a circle from a point outside it, the square of the distance to the circle along the tangent is equal to the product of the distances to the two points on the circle along the secant; in other words,

\(\displaystyle NA \cdot NB = (NT)^{2}\),

and, substituting, 

\(\displaystyle t(t+24) = 16 ^{2}\)

Distributing and writing in standard quadratic polynomial form,

\(\displaystyle t^{2} + 24t=256\)

\(\displaystyle t^{2} + 24t- 256 =256 - 256\)

\(\displaystyle t^{2} + 24t- 256 =0\)

We can factor the polynomial by looking for two integers with product \(\displaystyle -256\) and sum 24; through some trial and error, we find that these numbers are 32 and \(\displaystyle -8\), so we can write this as 

\(\displaystyle (t+32)(t- 8)= 0\)

By the Zero Product Principle, 

\(\displaystyle t+32 = 0\), in which case \(\displaystyle t = -32\) - impossible since \(\displaystyle t\) is a (positive) distance; or,

\(\displaystyle t-8 = 0\), in which case \(\displaystyle t = 8\) - the correct choice.

If two chords of a circle intersect inside the circle, the product of the lengths of the parts of each chord is the same. In other words,

\(\displaystyle t \cdot t = 8 \cdot 12\)

\(\displaystyle t ^{2} = 96\)

Solving for \(\displaystyle t\):

\(\displaystyle t = \sqrt{96 }\)

Simplifying the radical using the Product of Radicals Principle, and noting that the greatest perfect square factor of 96 is 16:

\(\displaystyle t = \sqrt{16 } \cdot \sqrt{6 } = 4 \sqrt{6 }\)

If a secant segment line and a tangent segment are constructed to a circle from a point outside it, the square of the length of the tangent is equal to the product of the distances to the two points on the circle intersected by the secant; in other words,

\(\displaystyle NA \cdot NB = (NT)^{2}\)

\(\displaystyle NA \cdot (NA+AB )= (NT)^{2}\)

Substituting:

\(\displaystyle 20 \cdot (20+t) = 25 ^{2}\)

Distributing, then solving for \(\displaystyle t\):

\(\displaystyle 20 \cdot 20+ 20 \cdot t= 6 25\)

\(\displaystyle 20t + 400 = 625\)

\(\displaystyle 20t + 400 - 400 = 625 - 400\)

\(\displaystyle 20t = 225\)

\(\displaystyle 20t \div 20 = 225 \div 20\)

\(\displaystyle t = 11.25\)