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Radius of Circle

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Radius of Circle

by dtweah » Wed Jun 03, 2009 3:18 pm
Let AB and CD be two chords of a circle that intersect at a point P. Suppose that AP = 4, PB = 6, CP = 2, PD = 12 and Angle APC = 90◦. What is the radius of the circle?

(a) 4√3
(b) 3√6
(c) 8
(d) 5√2
(e) √47
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Re: Radius of Circle

by Pranay » Thu Jun 04, 2009 3:25 am
dtweah wrote:Let AB and CD be two chords of a circle that intersect at a point P. Suppose that AP = 4, PB = 6, CP = 2, PD = 12 and Angle APC = 90◦. What is the radius of the circle?

(a) 4√3
(b) 3√6
(c) 8
(d) 5√2
(e) √47
I reached 2 root(5).

Please confirm the option d. is it 2√5 or 5√2 ??
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Re: Radius of Circle

by dtweah » Thu Jun 04, 2009 4:09 am
Pranay wrote:
dtweah wrote:Let AB and CD be two chords of a circle that intersect at a point P. Suppose that AP = 4, PB = 6, CP = 2, PD = 12 and Angle APC = 90◦. What is the radius of the circle?

(a) 4√3
(b) 3√6
(c) 8
(d) 5√2
(e) √47
I reached 2 root(5).

Please confirm the option d. is it 2√5 or 5√2 ??
All the answer choices are exactly as written above. Solve and post your reasoning. If there is error in your reasoning others may point it out.
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by PAB2706 » Thu Jun 04, 2009 4:27 am
Ans is D

Draw perpendiculars from center to both chords. Name points x nd y.
These perpendiculars bisect the resp chord.
OA is radius. You get AX=OX=5

Apply pythagorus thm to AOX you get OA=5rt2
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Re: Radius of Circle

by Pranay » Thu Jun 04, 2009 4:33 am
dtweah wrote:
Pranay wrote:
dtweah wrote:Let AB and CD be two chords of a circle that intersect at a point P. Suppose that AP = 4, PB = 6, CP = 2, PD = 12 and Angle APC = 90◦. What is the radius of the circle?

(a) 4√3
(b) 3√6
(c) 8
(d) 5√2
(e) √47
I reached 2 root(5).

Please confirm the option d. is it 2√5 or 5√2 ??
All the answer choices are exactly as written above. Solve and post your reasoning. If there is error in your reasoning others may point it out.
Hi dtweah,

Sorry for bothering you!!!

I missed out a basic out a basic point while drawing a conclusion .. so there an error in my computing .. !!

Anyways ... I am trying to solve it .. and will post if reach the conclusion. :)

Pranay
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Re: Radius of Circle

by Pranay » Mon Jun 08, 2009 3:17 am
Hi ..

The answer is D.

Using the Trignometry concepts, I could get that angle ADP is 45 degrees.
(Applying tan inverse to each of the right angled triangles ADP and BDP)

Using a theorm, the angles subtended by a chord at the edge of a circle is half that of subtended at the centre implies, angle AOB = 90 degrees where O is the centre of the circle.

Now drop a perpendicular from the O to chord AB, at point Z, to form a right angled triangle AZO.

=> AZ=BZ=5 (from question)
angle OZA = 90 degrees, angle ZOA = 45 degrees and angle OAZ = 45 degrees.

=> OA=OZ since, triangle OZA is an isosceles right angled triangle.

let OA = r (radius)

From Pythogaras theorm,

OZ^2 + AZ^2 = OA^2

Thus, by substituting the values we reach r = 5sqrt(2)

which is option D.

Hope the answer is right.

Also, please post the source of the question along with the OA. [:)]
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