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Bunuel

P is a point (not shown in the figure) in the plane containing O, R, and Q where O is the center of the circle and R and Q are points on the circle. If r is the radius of the circle, is P inside the circle?

(1) PR > PO
(2) PR + PQ = 5r



DS21262


Attachment:
The attachment 1.png is no longer available


Statement 1:
P must be further away from R than O, we can have the two cases showed in the attachment. Insufficient.

Statement 2:
While keeping P within the circle, PR is at most 2r. PQ is also at most 2r. Considering that, even if we add the maximums directly we cannot hit 5r. While keeping P inside the circle, the maximum of PR + PQ involves an optimization problem and it is certainly less than 4r. Therefore P must be outside the circle. Sufficient.

Ans: B
Attachments

File comment: For statement 1
Circle1.png
Circle1.png [ 25.13 KiB | Viewed 7873 times ]

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Bunuel

P is a point (not shown in the figure) in the plane containing O, R, and Q where O is the center of the circle and R and Q are points on the circle. If r is the radius of the circle, is P inside the circle?

(1) PR > PO
(2) PR + PQ = 5r



DS21262


Attachment:
1.png

(1) PR > PO

Consider point P in the top half of the circle somewhere. PR > PO.
Now move the point P vertically up outside the circle. Still PR > PO.
So P may be inside or outside the circle - not sufficient

(2) PR + PQ = 5r

If two points are inside or at the circle, maximum distance between them will be 2r (the diameter). For any points Q and R on the circle, the maximum sum of distance of P from Q and R will be 2 * 2r = 4r.
If the sum of the distances is 5r, P MUST be outside the circle.
Sufficient

Answer (B)
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Bunuel

P is a point (not shown in the figure) in the plane containing O, R, and Q where O is the center of the circle and R and Q are points on the circle. If r is the radius of the circle, is P inside the circle?

(1) PR > PO
(2) PR + PQ = 5r



DS21262


Attachment:
1.png

(1) One way to visualize this statement is that Point P is closer to Point O than Point R. This could certainly be possible inside the circle and outside the circle. INSUFFICIENT.

(2) \(PR + PQ = 5r\)

Given the radius = r
diameter = 2r

Since we're told PR + PQ = 5r, we can conclude with certainty that P is NOT in the circle because the diameter is 2r. SUFFICIENT.

Answer is B.
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