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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
44% (02:33) correct
56%
(02:18)
wrong
based on 50
sessions
History
Date
Time
Result
Not Attempted Yet
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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Attachment:
2020-06-19_1305.png [ 45.34 KiB | Viewed 4103 times ]
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Attachment:
2020-06-19_1305.png [ 45.64 KiB | Viewed 3556 times ]
AO *OB = OC*OD
2*6=3* OC
OC = 4
Suppose radius of the circle = r
\(4r^2 = OA^2 + OB^2 +OC^2 + OD^2\)
\(4r^2 = 2^2+6^2 + 4^2+3^2\)
\(r^2 = \frac{65}{4}\)
\(r = \frac{\sqrt{65}}{2}\)
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
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
Attachments
Untitled.png [ 54.39 KiB | Viewed 3634 times ]
Quote:
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)
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)
Hence answer is B
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)
Hence answer is B
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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 \(R=\frac{\sqrt{65}}{2}\)
Answer is (B)
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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 \(R=\frac{\sqrt{65}}{2}\)
Answer is (B)
Attachments
2020-06-19_1305.png [ 51.86 KiB | Viewed 3404 times ]
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GMAT 1: 740 Q50 V40
Posts: 13
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???
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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