Question

in the figure below, s is the center of the circle. suppose that lk = 24, kn = 12, jn = 2x - 2, and ns = 9.5. find the following.

Explanation:

Step1: Apply the power - of - a - point theorem

If two chords $JL$ and $QR$ intersect at a point $N$ inside the circle, then $JN\times NL=KN\times NR$. Since the perpendicular from the center of a circle to a chord bisects the chord, $NL = LK=24$ and $JN\times24=12\times12$. Also, $JN = 2x - 2$. So, $(2x - 2)\times24=12\times12$.

Step2: Solve the equation for $x$

First, simplify the right - hand side of the equation: $12\times12 = 144$. Then the equation becomes $(2x - 2)\times24=144$. Divide both sides by 24: $2x-2=\frac{144}{24}=6$. Add 2 to both sides: $2x=6 + 2=8$. Divide by 2: $x = 4$.

Step3: Find the radius of the circle

Since $NS = 9.5$, and the perpendicular from the center to a chord bisects the chord, consider the right - triangle formed by half of the chord and the line from the center to the mid - point of the chord. Let the radius of the circle be $r$. In right - triangle $SNP$, we know that the circle's radius $r$ is related to the segments of the chords. Since the perpendicular from the center to a chord bisects the chord, and we know the relationship between the chords and the center. The radius of the circle $r$ is the distance from $S$ to any point on the circle. In right - triangle $SNP$, we note that the radius of the circle is the same as the distance from $S$ to any point on the circle. Since $NS = 9.5$, and considering the symmetry of the circle and the perpendiculars from the center to the chords, $PS=NS = 9.5$.

Answer:

$x = 4$
$PS = 9.5$