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

Question

Competition Mode Question https://gmatclub.com/forum/download/file.php?id=55431 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 Are You Up For the Challenge: 700 Level Questions 2020-06-19_1305.png

nick1816 · Answer

2020-06-19_1305.png 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}

Lipun · Answer

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) =  /2

Ans: B

exc4libur · Answer

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+(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- )^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)

dtox194 · Answer

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

firas92 · Answer

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)

vtexcelgmat · Answer

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

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

Prep Toolkit

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