Question

what is the length of line segment sv? 8 units 6 units 16 units 12 units

Explanation:

Step1: Apply the secant - secant rule

If two secants are drawn to a circle from an external point, then $(y - 2)(y - 2+(y + 4))=6\times(6 + 8)$.

Step2: Simplify the left - hand side and right - hand side

First, simplify the left - hand side: $(y - 2)(2y+2)=2y^{2}+2y-4y - 4=2y^{2}-2y - 4$. The right - hand side is $6\times14 = 84$. So, $2y^{2}-2y - 4=84$, which simplifies to $y^{2}-y - 44 = 0$.

Step3: Solve the quadratic equation

Using the quadratic formula $y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $y^{2}-y - 44=0$ where $a = 1$, $b=-1$, and $c=-44$. $y=\frac{1\pm\sqrt{(-1)^{2}-4\times1\times(-44)}}{2\times1}=\frac{1\pm\sqrt{1 + 176}}{2}=\frac{1\pm\sqrt{177}}{2}$. But we can also use the secant - secant relationship in another way.
If we consider the power of a point $V$ with respect to the circle: $(SV)(SW)=(UV)(TV)$. Let's assume we use the property that if we have two secants $SVT$ and $UVW$ from an external point $V$ to the circle.
We know that $(y + 4)(y + 4+(y - 2))=(8)(8 + 6)$.
$(y + 4)(2y + 2)=112$.
$2y^{2}+2y+8y + 8=112$.
$2y^{2}+10y-104 = 0$.
$y^{2}+5y - 52=0$.
$(y + 8)(y - 3)=0$.
So, $y = 3$ or $y=-8$. Since length cannot be negative, $y = 3$.

Step4: Calculate the length of SV

$SV=y + 4+6$. Substitute $y = 3$ into the expression, $SV=3 + 4+6=16$ units.

Answer:

16 units