BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Clarification

Expert replies
by vladmire » Sun Dec 07, 2008 12:47 pm
In the xy-plane, each point on circle K has nonnegative coordinates and the center of K is the point (4,7). What is the maximum possible area of K?
4pi
9pi
16pi
28pi
49pi

The way I went about solving this was to draw a xy-grid so the circle was in quadrant I. Because all of the points on the circle are positive. Then the radius is 4, because the center of circle is 4 away from the x axis. Area of a circle is a=pir^2 so 4^2=16 or 16pi cannot be any larger right cause the points must all be positive.
Join the discussion
Source: — Problem Solving |

by kris610 » Sun Dec 07, 2008 2:16 pm
I'll go with 16pi. Draw perpendiculars from the center to the x and y axes--then take whichever is less.
Join the discussion