1
$\begingroup$

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!

$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

Extend $OM$ to $N$ on the circle, and let the tangent at $N$ meet $SR$ extended at $L$.

Then by Euclid, Elements III, 36$$NL^2=LR\cdot LS$$length of OKAnd 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$$OK=\sqrt{3^2+3^2}=\sqrt{18}=3\sqrt2$$

$\endgroup$
0
$\begingroup$

if you regard the chords as the axes of a rectangular co-ordinate system, with origin K, then you have shown that the coordinates of the center of the circle, O, must be $(3,3)$. So the distance OK is $3 \sqrt{2}$

$\endgroup$
2
  • $\begingroup$ Thanks for answering @davidholden but I do not get how O= (3,3) and is it possible without using the coordinate system? $\endgroup$ Commented Oct 14, 2018 at 5:36
  • $\begingroup$ your Pythagoras argument shows that the distance from the circle center O to any chord $c$ of length 8 cm must be 3 cm. but this distance is just the co-ordinate of O measured from K along a chord perpendicular to $c$. draw a diagram and it should be clear $\endgroup$ Commented Oct 14, 2018 at 5:41

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.