A circle lies in an X-Y plane, with its center (1, -3)

[jsl]

A circle lies in an X-Y plane, with its center at (1, -3). If the distance between the X-intercepts of the circle is 8 units, what is the area of the circle?

9Pi
16Pi
25Pi
36Pi
64Pi

(dodgy question)... answer is 25Pi anyhow

[earth@work]

My answer is no where close to any of these options...dnt know where m getting wrong.
X-intercepts will be (5,0) and (-3,0)
dropping a perpendicular to x-axis, we get length of perpend = 3 coz (1.-3) is the center of the circle, the other side of triangle formed (on x-axis)=4 (=8/2)
Radius will be equal to the hypot. by applying pyth theorem
4^2+3^2=r^2
r=5
area=5x5x22/7=78 approx..
what m i doing wrong?
sorry cudn't help jsl.

[Ian Stewart]

A circle lies in an X-Y plane, with its center at (1, -3). If the distance between the X-intercepts of the circle is 8 units, what is the area of the circle?

9
16
25
36
64

None of the answer choices above is correct, clearly; there should be a Pi in each, presumably. I know the source of the question, and it's their mistake, not jsl's. In any case, if the x-intercepts are 8 apart, by symmetry they must be at (-3, 0) and (5, 0); that is, the x-co-ordinates of the intercepts must be 4 to the left and right of x=1, the x-co-ordinate of the centre of the circle. This is easier to see if you draw a diagram in the co-ordinate plane. Once you know these points on the circle, you can find that the distance to the centre is 5, so the circle's radius is 5. The area is then Pi*(5)^2 = 25*Pi.

[earth@work]

thanks Ian... i was thinking that i m making some mistake in calculation or some error in my method

[gmat009]

A circle lies in an X-Y plane, with its center at (1, -3). If the distance between the X-intercepts of the circle is 8 units, what is the area of the circle?

9
16
25
36
64

None of the answer choices above is correct, clearly; there should be a Pi in each, presumably. I know the source of the question, and it's their mistake, not jsl's. In any case, if the x-intercepts are 8 apart, by symmetry they must be at (-3, 0) and (5, 0); that is, the x-co-ordinates of the intercepts must be 4 to the left and right of x=1, the x-co-ordinate of the centre of the circle. This is easier to see if you draw a diagram in the co-ordinate plane. Once you know these points on the circle, you can find that the distance to the centre is 5, so the circle's radius is 5. The area is then Pi*(5)^2 = 25*Pi.

Just one confusion Ian,
How we got this radius 5. IMO radius shud be 4.
Circle has centre at (1,-3) so radius shud be 4 instead of 5.
Plz. explain this.......

[Ian Stewart]

A circle lies in an X-Y plane, with its center at (1, -3). If the distance between the X-intercepts of the circle is 8 units, what is the area of the circle?

9
16
25
36
64

None of the answer choices above is correct, clearly; there should be a Pi in each, presumably. I know the source of the question, and it's their mistake, not jsl's. In any case, if the x-intercepts are 8 apart, by symmetry they must be at (-3, 0) and (5, 0); that is, the x-co-ordinates of the intercepts must be 4 to the left and right of x=1, the x-co-ordinate of the centre of the circle. This is easier to see if you draw a diagram in the co-ordinate plane. Once you know these points on the circle, you can find that the distance to the centre is 5, so the circle's radius is 5. The area is then Pi*(5)^2 = 25*Pi.

Just one confusion Ian,
How we got this radius 5. IMO radius shud be 4.
Circle has centre at (1,-3) so radius shud be 4 instead of 5.
Plz. explain this.......

The center is at (1, -3), and (5,0) is a point on the circle, The distance from (1, -3) and (5,0) is 5, not 4 (draw the picture; you should get a 3-4-5 triangle).

[dmateer25]

Hope this image helps

[gmat009]

A circle lies in an X-Y plane, with its center at (1, -3). If the distance between the X-intercepts of the circle is 8 units, what is the area of the circle?

9
16
25
36
64

None of the answer choices above is correct, clearly; there should be a Pi in each, presumably. I know the source of the question, and it's their mistake, not jsl's. In any case, if the x-intercepts are 8 apart, by symmetry they must be at (-3, 0) and (5, 0); that is, the x-co-ordinates of the intercepts must be 4 to the left and right of x=1, the x-co-ordinate of the centre of the circle. This is easier to see if you draw a diagram in the co-ordinate plane. Once you know these points on the circle, you can find that the distance to the centre is 5, so the circle's radius is 5. The area is then Pi*(5)^2 = 25*Pi.

Just one confusion Ian,
How we got this radius 5. IMO radius shud be 4.
Circle has centre at (1,-3) so radius shud be 4 instead of 5.
Plz. explain this.......

The center is at (1, -3), and (5,0) is a point on the circle, The distance from (1, -3) and (5,0) is 5, not 4 (draw the picture; you should get a 3-4-5 triangle).

Got it.....
Thanks Ian

[gmat009]

Hope this image helps

Great......
Thanks