guys, I have worked through this problem although I found possible answers but I am still unsure.
Question: PQ and RS are two perpendicular chords of a circle, centred at O of radius 5cm, intersecting at K. Each chord is 8cm long, find the length of OK.
My workings:
RS=8, PQ=8 PM(for midpoint) = MQ (line from centre to mid-pt chord)
$PO^2=PM^2 + MO^2$ (Phytagoras), $5^2=4^2 + MO^2$, $3cm$ = MO
Now, this is the part where I am stuck, but I do know that in order to find OK I need Pythagoras. Please help thanks!
Best Answer
Extend $OM$ to $N$ on the circle, and let the tangent at $N$ meet $SR$ extended at $L$.
Then by Euclid, Elements III, 36$$NL^2=LR\cdot LS$$
And since$$RS=8$$and$$LR=\frac{10-8}{2}=1$$then$$LN^2=1\cdot 9$$and$$LN=KM=3$$But equal chords in a circle are equidistant from the center. Therefore$$OM=KM=3$$and by Pythagoras$$OK=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt2$$