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?

Re: Finding radius of circle with 2 Perpendicular Chords

Quote:

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 $\displaystyle |\overline{VP}|$ 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.

Re: Finding radius of circle with 2 Perpendicular Chords

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

Re: Finding radius of circle with 2 Perpendicular Chords

Quote:

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.

Re: Finding radius of circle with 2 Perpendicular Chords

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

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

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