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# In the diagram above, what is the length of the radius of the circle?

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Math Expert
Joined: 02 Sep 2009
Posts: 65785
In the diagram above, what is the length of the radius of the circle?  [#permalink]

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19 Jun 2020, 01:07
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75% (hard)

Question Stats:

33% (02:31) correct 67% (02:13) wrong based on 36 sessions

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Competition Mode Question

In the diagram above, what is the length of the radius of the circle?

A. 4

B. $$\frac{\sqrt{65}}{2}$$

C. 4.5

D. 5

E. 6

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Re: In the diagram above, what is the length of the radius of the circle?  [#permalink]

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19 Jun 2020, 02:09
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The products of the lengths of the line segments on each intersecting chord are equal.

AO *OB = OC*OD
2*6=3* OC

OC = 4

Suppose radius of the circle = r

$$4r^2 = OA^2 + OB^2 +OC^2 + OD^2$$

$$4r^2 = 2^2+6^2 + 4^2+3^2$$

$$r^2 = \frac{65}{4}$$

$$r = \frac{\sqrt{65}}{2}$$
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Re: In the diagram above, what is the length of the radius of the circle?  [#permalink]

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19 Jun 2020, 03:57
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Triangles AZC and DZB are similar. So, ZC = (2/3)*6 = 4

AB and CD are two chords of a circle. Perpendicular from the centre of a circle bisects the chord.

So, x co-ordinate of X is 4 and y co-ordinate of Y is 0.5. => O = (4,0.5)

Radius = OA = root(4^2 + 0.5^2) = [root(65)]/2

Ans: B
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Re: In the diagram above, what is the length of the radius of the circle?  [#permalink]

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19 Jun 2020, 04:55
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Quote:
In the diagram above, what is the length of the radius of the circle?

A. 4
B. 65/√2
C. 4.5
D. 5
E. 6

distance between two points: d^2=(x1-x2)^2+(y1-y2)^2
circle center coordinates: (a,b)
two chords are x and y axis
r^2=(a-6)^2+(b-0)^2,
r^2=a^2+36-12a+b^2;
r^2=(a-[-2])^2+(b-0)^2,
r^2=a^2+4+4a+b^2;
a^2+36-12a+b^2=a^2+4+4a+b^2;
a=32/16=2

r^2=(a-0)^2+(b-[-3])^2,
r^2=(2)^2+(b+3)^2,
r^2=16+b^2+9+6b

a^2+4+4a+b^2=16+b^2+9+6b,
16+4+8+b^2=25+b^2+6b,
6b=28-25, b=3/6=1/2

r^2=(2-6)^2+(0.5-0)^2,
r^2=16+0.25, r=V(16.25),
r=V(16+1/4), r=V(65/4)=V65/2

Ans (B)
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Re: In the diagram above, what is the length of the radius of the circle?  [#permalink]

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19 Jun 2020, 13:45
1
1
Let perpendicular chords meet at P, then CP = 6*2/3 = 4
Assume center of circle is at O. drop perpendiculars from center O onto both the chords. it will bisect the chords CD and AB respectively at midpoints M and N respectively.

Therefore CM = 7/2 and BN = 4 (consequently, NP=2)

In right triangle OMC, r^2= CM^2 + OM^2 = CM^2+NP^2
=> r^2 = (7/2)^2 + 2^2
=> r^2 = 65/4
r= sqrt (65/4)

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Re: In the diagram above, what is the length of the radius of the circle?  [#permalink]

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20 Jun 2020, 02:04
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1
Since $$AB$$ and $$CD$$ are intersecting chords, $$AE*EB=CE*ED$$

Or $$CE=\frac{2*6}{3}=4$$

$$EB=6, FB=4$$ and so $$EF=EB-FB=6-2=2$$

Similarly,

$$DG=\frac{7}{2}, DE=3$$ and so $$GE=DG-DE=\frac{7}{2}-3=\frac{1}{2}$$

Now from right triangle $$AOF$$,

$$AF^2+OF^2=R^2$$ where $$R$$ is the radius of the circle

$$AF^2+GE^2=R^2$$ (Since $$OF=GE$$)

$$4^2+(\frac{1}{2})^2=R^2$$,

Solving, we get $$R=\frac{\sqrt{65}}{2}$$

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Re: In the diagram above, what is the length of the radius of the circle?  [#permalink]

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20 Jun 2020, 11:09
1
ANSWER is B $$\sqrt{65}/2$$
BUT THIS TOOK TOO MUCH TIME AND EQUATION SOLVING.

Consider the image as a Graph used for Coordinate Geometry.
Use the equation of circle (x - a)^2 + (y - b)^2 = r^2
Note down the three points from the Diagram.
(-2,0)
(0,-3)
(6,0)
You will get following three equations:
(-2 - a)^2 + (0 - b)^2 = r^2
(0 - a)^2 + (-3 - b)^2 = r^2
(6 - a)^2 + (0 - b)^2 = r^2
Solving the equations will give the following values:
a= 2
b= ½
r= $$\sqrt{65}/2$$

Does anyone have a WAY that does not charge so much time???
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Re: In the diagram above, what is the length of the radius of the circle?   [#permalink] 20 Jun 2020, 11:09