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M23 q 27 : Retired Discussions [Locked]

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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?

* 36\pi * 45\pi * 64\pi * 81\pi * 100\pi

Can someone please explain this question stem with figure?

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?

* 36\pi * 45\pi * 64\pi * 81\pi * 100\pi

Can someone please explain this question stem with figure?

Below is a diagram:

Attachment:

Circle.png [ 11.15 KiB | Viewed 1486 times ]

As you can see the radius of the circle is a hypotenuse of a right triangle with the sides equal to 6 and 16/2=8 (6-8-10 right triangle), so radius=hypotenuse=10. The area is \pi{r^2}=100\pi.

The center of the circle is at point (0, 6)[/m]. If the [#permalink]
13 Dec 2012, 02:01

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?

A. 36\pi B. 45\pi C. 64\pi D. 81\pi E. 100\pi

Last edited by Bunuel on 13 Dec 2012, 02:05, edited 1 time in total.

Re: The center of the circle is at point (0, 6)[/m]. If the [#permalink]
13 Dec 2012, 02:09

Expert's post

arnijon90 wrote:

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?

A. 36\pi B. 45\pi C. 64\pi D. 81\pi E. 100\pi

Merging similar topics. Please refer to the solutions above. _________________

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?

* 36\pi * 45\pi * 64\pi * 81\pi * 100\pi

Can someone please explain this question stem with figure?

Below is a diagram:

Attachment:

Circle.png

As you can see the radius of the circle is a hypotenuse of a right triangle with the sides equal to 6 and 16/2=8 (6-8-10 right triangle), so radius=hypotenuse=10. The area is \pi{r^2}=100\pi.

Answer: E.

Hope it helps.

M taking abt y intercept which is 16 and x intercept -4?How did u draw this ! I understood that it will cross 8 and -8 on both side of x axis. _________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

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?

* 36\pi * 45\pi * 64\pi * 81\pi * 100\pi

Can someone please explain this question stem with figure?

Below is a diagram:

Attachment:

Circle.png

As you can see the radius of the circle is a hypotenuse of a right triangle with the sides equal to 6 and 16/2=8 (6-8-10 right triangle), so radius=hypotenuse=10. The area is \pi{r^2}=100\pi.

Answer: E.

Hope it helps.

M taking abt y intercept which is 16 and x intercept -4?How did u draw this ! I understood that it will cross 8 and -8 on both side of x axis.

Sorry but do not follow you. What do you mean by "how did you draw it"?

If you understand that x intercepts are at -8 and 8, then you can get the radius as shown in my post. Drawing is just to illustrate. _________________

Ohk Bunuel dear .. Finally I got that now.. hypotenuse is 10 so from 6 to -4 distance is 10 and from 6 to 16 distance is 10..Its clear now _________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

How will I understand without drawing that the two points are (0, -8) and (0, 8)? Why not (0, -7) and (0, 9)? Please make it clear.

The center of the circle is on y-axis, so it's symmetric around it. We know that the the distance between the two points where the circle intersects the x-axis is 16, hence half of it (8 units) must be to the left of 0 and the remaining half (another 8 units) to the right of 0.