Question

in the figure below, s is the center of the circle. suppose that $jk = 16$, $mp = 8$, $lp = 2x + 4$, and $ps = 3$. find the following.
$x = \square$
$ns = \square$

Explanation:

Step1: Find radius via chord MR

First, note that \(SP \perp MR\), so \(MR = 2 \times MP = 2 \times 8 = 16\). Let the radius of the circle be \(r\). In right triangle \(SPM\):
\(r^2 = MP^2 + PS^2\)
\(r^2 = 8^2 + 3^2 = 64 + 9 = 73\)
So \(r = \sqrt{73}\)

Step2: Find radius via chord JL

Since \(SN \perp JK\), \(JN = \frac{JK}{2} = \frac{16}{2} = 8\). In right triangle \(SNJ\):
\(r^2 = JN^2 + NS^2\)
\(73 = 8^2 + NS^2\)
\(73 = 64 + NS^2\)
\(NS^2 = 73 - 64 = 9\)
\(NS = 3\)

Step3: Solve for x using chord JL

Chord \(JL = 2 \times LP = 2(2x+4)\). In right triangle \(SPL\):
\(r^2 = LP^2 + PS^2\)
\(73 = (2x+4)^2 + 3^2\)
\(73 = (2x+4)^2 + 9\)
\((2x+4)^2 = 73 - 9 = 64\)
Take square root: \(2x+4 = 8\) (since length positive)
\(2x = 8 - 4 = 4\)
\(x = \frac{4}{2} = 2\)

Answer:

\(x = 2\)
\(NS = 3\)