Coordinate geometry

the area of a circly is pi(r)^2 - so in circle area problems we need to find the radius. here we are given (4,7) as the center and the fact that there are no negative points on the circle, which means the circle must remain in the first quadrant. So we have to find the maximum radius in the first quadrant - this is 4 becuase if the radius were any bigger the circle would cross over to negative x values. so becuase the radius must be 4 the area must be 16 pi.

each point on a circle k has non negative coordinates and center of k is the point (4,7).
the above part implies that max radius is 4, any thing above 4 will make the circle pass through -ve quardant
therefore max area is 16pi

tpr-becky wrote:the area of a circly is pi(r)^2 - so in circle area problems we need to find the radius. here we are given (4,7) as the center and the fact that there are no negative points on the circle, which means the circle must remain in the first quadrant. So we have to find the maximum radius in the first quadrant - this is 4 becuase if the radius were any bigger the circle would cross over to negative x values. so becuase the radius must be 4 the area must be 16 pi.

Becky,

I can't understand this "becuase if the radius were any bigger the circle would cross over to negative x values."


Can you please explain?

Because the X value on the number line is 4 if the radius gets any bigger than 4 the circle will cross over into negative x values - eg. if the radius is 1 then the x values for the edges of the circle are 3 and 5 - if the radius is 5 then the values are -1 and 9 - which breaks the rules. therefore the farthest the left edge of the circle can go is to 0 - which means the radius must be 4.

Got it..Thanks

The area of a circle is determined by the radius, so this question can be thought of as "What is the longest radius possible if all the points of the circles must have non-negative coordinates?"

non-negative coordinates are only possible if the circle stays entirely in the top right quadrant. The center of the circle is at (4,7) so it is 4 units to the right of the vertical axis (y-axis) and 7 units above the horizontal axis (x-axis). As the circle gets larger and larger, the first points to fall outside of the top right quadrant will occur when the circle crosses the vertical axis.

Therefore to keep all points non-negative, we must be sure that the radius is not longer than the distance from the center to the vertical axis. The radius cannot be greater than 4, which means that the area cannot be greater than 16pi (because area is pi*radius^2)

If you need illustrations to better understand this, have a look at GMATPrep question 1525. You can practice similar questions if you have access to the Solutions Engine drill generator by selecting topic="Geometry, Coordinate" and difficulty="500-600"

Good luck,
-Patrick

what is the drill generator how much does it cost
where to download it from

Patrick_GMATFix wrote:The area of a circle is determined by the radius, so this question can be thought of as "What is the longest radius possible if all the points of the circles must have non-negative coordinates?"

non-negative coordinates are only possible if the circle stays entirely in the top right quadrant. The center of the circle is at (4,7) so it is 4 units to the right of the vertical axis (y-axis) and 7 units above the horizontal axis (x-axis). As the circle gets larger and larger, the first points to fall outside of the top right quadrant will occur when the circle crosses the vertical axis.

Therefore to keep all points non-negative, we must be sure that the radius is not longer than the distance from the center to the vertical axis. The radius cannot be greater than 4, which means that the area cannot be greater than 16pi (because area is pi*radius^2)

If you need illustrations to better understand this, have a look at GMATPrep question 1525. You can practice similar questions if you have access to the Solutions Engine drill generator by selecting topic="Geometry, Coordinate" and difficulty="500-600"

Good luck,
-Patrick