Two chords KV and QR of circle O are perpendicular at point P, with PQ=6 and PR=8 If radius of circle is sq root of 65. find KP and PV.
The answer is KP=12 and PV=4
My ques is how did they get KP? i understand how they get PV since QP*PR=KP*PV
But i dont know what to do with the radius to get KP?
The centre lies on the perpendicular bisector of QR,
which is parallel to KV
and lies 1 unit below KV (from the sketch).
Therefore Pythagoras' Theorem applied to the right-angled triangle OKM or OVM,
where O is the centre and M is the midpoint of KV yields Earboth's 2nd equation.
You could also have answered exclusively with Pythagoras' Theorem.
Label the circle centre O, the midpoint of KV = M and the midpoint of QR = N.
Then the circle centreline containing O and N lies 1 unit below [KV]
as it is 7 units from Q and 7 units from R.
M is the midpoint of [KV]
Label the point of intersection of [QR] and the horizontal centreline as S.
Then
Then