# Finding radius of circle with 2 Perpendicular Chords

• Jun 15th 2011, 09:33 PM
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?
• Jun 15th 2011, 10:21 PM
earboth
Re: Finding radius of circle with 2 Perpendicular Chords
Quote:

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 $\displaystyle |\overline{VP}|$ and y denotes $\displaystyle |\overline{PK}|$.
According to the intersecting chord theorem (google for it!)[1st equation] and Pythagorean theorem[2nd equation] you'll get the system of equations:

$\displaystyle \left|\begin{array}{rcl}x \cdot y &=& 6 \cdot 8 \\ \left(\dfrac{x+y}{2}\right)^2+1^2 &=& 65 \end{array}\right.$

3. Solve for x and y.
• Jun 16th 2011, 05:48 AM
Re: Finding radius of circle with 2 Perpendicular Chords
I understand the intersecting chord theorem. But how did you derive the 2nd equation?
• Jun 16th 2011, 11:27 AM
Re: Finding radius of circle with 2 Perpendicular Chords
Quote:

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.
• Jun 16th 2011, 07:10 PM
Re: Finding radius of circle with 2 Perpendicular Chords
Thanks. i get it now. i was making it way to complicated
• Jun 17th 2011, 02:29 AM
Re: Finding radius of circle with 2 Perpendicular Chords
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]

$\displaystyle \Rightarrow\ |KM|^2+1=65\Rightarrow\ |KM|=8$

Label the point of intersection of [QR] and the horizontal centreline as S.

Then

$\displaystyle |OS|^2+7^2=65\Rightarrow\ |OS|=4$

Then

$\displaystyle |KV|=2|KM|,\;\;\;|KP|=|KM|+|OS|,\;\;\;|PV|=|KM|-|OS|$