Thread: Finding radius of circle with 2 Perpendicular Chords

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?

Originally Posted by Aquameatwad

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?

1. Draw a sketch.

2. Let x denote and y denotes .
According to the intersecting chord theorem (google for it!)[1st equation] and Pythagorean theorem[2nd equation] you'll get the system of equations:



3. Solve for x and y.

I understand the intersecting chord theorem. But how did you derive the 2nd equation?

Originally Posted by Aquameatwad

I understand the intersecting chord theorem. But how did you derive the 2nd equation?

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.

Thanks. i get it now. i was making it way to complicated

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