If A and C are points in a plane, C is the center of circle

(1) 2πr=xπ
2r=x-----Dia of circle.
if c is centre and AC=x then point A must lie outside of circle.
(2)πr^2=xπ
r^2=x
x=2 it means AC=2 and radius of circle is Sq.root2 then A may lie inside or outside the circle since dia. of circle is 2*sq.root2....insuff...
Ans A

Hi Bunuel

Do we have point circle concept in Gmat?
if so, From (1) 2r=x=0 could also be a case and so point lie on circle. And this may become insufficient.

From (2) It's anyways eliminated even without the concept of point circle

From (1) & (2) we still have two cases so It may lead to "E"


From OG:
This is how they defined Line segment: ( It was not mentioned if P & Q should be distinct.)
The part of the line from P to Q is called a line segment. P and Q are the endpoints of the segment. The notation PQ is used to denote line segment PQ and PQ is used to denote the length of the segment.

This is how circe is defined: (The set can also have just one element?)
A circle is a set of points in a plane that are all located the same distance from a fixed point.

Please clear my confusion

Thanks in advance

Obs.: xπ is measured in unit of length, while areas are measured in unit of length squared. That´s why we modified statement (2) accordingly.

\({\text{dist}}\left( {A,C} \right)\,\,\, = \,\,x\,\,\,\,\mathop > \limits^? \,\,\,r\)

\(\left( 1 \right)\,\,\,\,2\pi r = \pi x\,\,\,\,\mathop \Rightarrow \limits^{:\,\,\pi \,} \,\,\,\,\,x = 2r\,\,\mathop > \limits^{r\,\, > \,0} r\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)

\(\left( 2 \right)\,\,\,\pi {r^2} = x\pi \,\,\,\,\mathop \Rightarrow \limits^{:\,\,\pi \,} \,\,\,\,\,x = {r^2}\,\,\,\left\{ \begin{gathered}\\
\,{\text{Take}}\,\,r = 2\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 4 > 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\\\
\,{\text{Take}}\,\,r = 0.5\,\,\,\,\, \Rightarrow \,\,\,\,\,x = 0.25 < 0.5\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{N}}{\text{O}}} \right\rangle \hfill \\ \\
\end{gathered} \right.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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