Question

use the circle below to find each measure given that hk = 30 and pm = 8.
ph = type your answer...
pn = type your answer...
mg = type your answer...

Explanation:

Step1: Find the length of \(PH\)

Since \(HK\) is a chord and \(PH\) is the radius. If \(HK = 30\), and \(PH\) is the radius. Let's assume the circle has center \(P\). The length of \(PH\) is the radius. If we consider the right - triangle formed (e.g., when a perpendicular from the center to a chord bisects the chord). But if we assume \(HK\) is a diameter (if the problem has no other constraints about the position of \(H\) and \(K\) relative to \(P\) in a wrong way of understanding, but if we use the property that \(PH\) is a radius. If we assume \(HK\) is a chord and use the Pythagorean theorem in the right - triangle \(PNK\) (where \(NK=\frac{HK}{2} = 15\), \(PH\) is the radius \(r\), \(PN\) is the distance from the center to the chord \(HK\)). But if we assume \(HK\) is a diameter (a wrong initial assumption but if no other info), \(PH=\frac{HK}{2}=15\). Wait, no, if \(HK\) is a chord, and \(PH\) is radius. Let's use the property: If a line from the center of a circle \(P\) is perpendicular to a chord \(HK\) at \(N\), then \(HN=\frac{HK}{2}=15\). By the Pythagorean theorem in right - triangle \(PNH\), if \(PH\) is the radius \(r\), \(PN\) is the distance from the center to the chord. But if we consider another chord \(GL\) (assuming \(PM\) is the distance from the center \(P\) to chord \(GL\) and chords equidistant from the center are equal. Wait, no, if \(PM = 8\) (distance from center \(P\) to chord \(GL\)) and assume \(HK\) and \(GL\) are related. Wait, no, \(PH\) is radius. If we assume \(HK\) is a chord and use the formula \(r^{2}=d^{2}+(\frac{l}{2})^{2}\) (where \(r\) is radius, \(d\) is distance from center to chord, \(l\) is chord length). But if \(PM = 8\) (distance from center to another chord \(GL\)) and assume \(HK\) and \(GL\) are congruent? No. Wait, no, \(PH\) is radius. If \(HK\) is a chord and \(PN\) is the distance from center \(P\) to \(HK\). Wait, no, \(PH\) is radius. If \(HK = 30\) is a chord and assume \(PN\) is the distance from \(P\) to \(HK\). But if \(PM = 8\) (distance from \(P\) to another chord \(GL\)) and assume \(GL = HK\) (chords equidistant from center are equal). No, wrong. Wait, \(PH\) is radius. If \(HK\) is a diameter, \(PH=\frac{HK}{2}=15\).

Step2: Find \(PN\)

Assume \(HK\) is a chord. Using the Pythagorean theorem in right - triangle \(PNH\) (where \(HN=\frac{HK}{2} = 15\), \(PH\) is radius. If \(PH = 17\) (from \(r^{2}=d^{2}+(\frac{l}{2})^{2}\), assume \(PM\) is wrong info used wrong. Wait, no, if \(PM = 8\) (distance from center to chord \(GL\)) and assume \(GL\) and \(HK\) are such that \(PH\) (radius) is calculated as follows: Let \(PH=r\), for chord \(GL\), if \(PM = 8\) (distance from center \(P\) to \(GL\)) and assume \(GL = HK = 30\) (chords equidistant from center are equal). Then \(r^{2}=8^{2}+15^{2}\) (since \(\frac{GL}{2}=\frac{HK}{2} = 15\)). So \(r=\sqrt{64 + 225}=\sqrt{289}=17\). Then \(PH = 17\). Then for \(PN\), if we assume \(PN = PM\) (if chords \(GL\) and \(HK\) are congruent and equidistant from center, but no, if \(PH = 17\), \(HN = 15\), then \(PN=\sqrt{PH^{2}-HN^{2}}=\sqrt{17^{2}-15^{2}}=\sqrt{(17 + 15)(17 - 15)}=\sqrt{32\times2}=\sqrt{64}=8\).

Step3: Find \(MG\)

Since \(PM = 8\) (distance from center \(P\) to chord \(GL\)), and \(GL=HK = 30\) (if chords equidistant from center are equal), then \(\frac{GL}{2}=15\). In right - triangle \(PMG\), \(MG=\sqrt{PG^{2}-PM^{2}}\). Since \(PG = PH = 17\) (radii of the same circle), \(MG=\sqrt{17^{2}-8^{2}}=\sqrt{(17 + 8)(17 - 8)}=\sqrt{25\times9}=15\).

Answer:

\(PH = 17\)
\(PN = 8\)
\(MG = 15\)