The center of the circle is at point (0,6). If the distance between the two points
where the circle intersects the xaxis is 16, what is the area of the circle?
Area
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I don't think the answer is 36pi it is only the centre. To find the area , we have to determine the radius, which according to the Q stem can be derived from the two points at which the circle intersects at x axis, say (x,0) and (x,0). The distance given between the two points is 16,
sqrt((2x)^2)=16 ===>x=2 or 2
Therefore, the two points are (2,0) and (2,0). The radius will be then distance between (0,6)and (2,0)
or (0,6) and (2,0) which gives sqrt(40). The area will then be 40pi.
But I am not sure whether this is the correct answer or the way to derive it. Need confirmation. Thanks.
sqrt((2x)^2)=16 ===>x=2 or 2
Therefore, the two points are (2,0) and (2,0). The radius will be then distance between (0,6)and (2,0)
or (0,6) and (2,0) which gives sqrt(40). The area will then be 40pi.
But I am not sure whether this is the correct answer or the way to derive it. Need confirmation. Thanks.
 shovan85
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The center is at (0,6). The circle formed around this center has to be symmetrical against the Yaxis.
The distance between the two points intersecting Xaxis is 16. So, we can say the length of the chord is 16.
The distance of any of the points from origin (0,0) is 8 as the circle is symmetric on Yaxis. (i.e the two intersecting points are (8,0) and (8,0))
Now the radius = sqrt(6^2 + 8^2) = 10
Thus, the area = 100 * pie
The distance between the two points intersecting Xaxis is 16. So, we can say the length of the chord is 16.
The distance of any of the points from origin (0,0) is 8 as the circle is symmetric on Yaxis. (i.e the two intersecting points are (8,0) and (8,0))
Now the radius = sqrt(6^2 + 8^2) = 10
Thus, the area = 100 * pie
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if you draw a triangle using center O and two points of intersection call it A and B.
then you get a isoceles triangle since both OA and OB are radius so OA=OB.
And since center is 0,6 which means  Y axis divides AB symterically at 90 degress using pythagorous theorum (hyp) OA=sqrt(6^2 + 8^2) = 10
Area= Pie *10^2=100pie
then you get a isoceles triangle since both OA and OB are radius so OA=OB.
And since center is 0,6 which means  Y axis divides AB symterically at 90 degress using pythagorous theorum (hyp) OA=sqrt(6^2 + 8^2) = 10
Area= Pie *10^2=100pie

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 amit2k9
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doing a bit of construction we have triangles with 6,8 and r as sides.
thus r=10 hence A = pi* 100.
thus r=10 hence A = pi* 100.
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Hi,
You are calculating r^2 = 6^2 + 16^2 instead of r^2 = 6^2 + 8^2.
Cheers!
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