The center of the circle is at point (0,6). If the distance between the two points
where the circle intersects the x-axis 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.
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The center is at (0,6). The circle formed around this center has to be symmetrical against the Y-axis.
The distance between the two points intersecting X-axis 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 Y-axis. (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 X-axis 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 Y-axis. (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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- mayank82
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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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