Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

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

Shortest Distance between a Circle and a Line

Expert replies
by ManuGMAT » Mon Jun 29, 2009 6:51 pm
What is the least possible distance between a point on the circle X^2+Y^2=1 and a point on the line Y=3/4X-3 ?

A. 1.4
B. 2^1/2
C. 1.7
D 3^1/2
E. 2.0

Please explain the answer.

OA is A
Join the discussion
Source: — Problem Solving |

by ssmiles08 » Mon Jun 29, 2009 7:53 pm
I sort of got the answer...but my method was somewhat crude...maybe some one has a better answer for you.

First I figured out the x and y intercepts of the line. (4,0) and (-3,0). this told me right away that there is some sort of a 3-4-5 triangle involved with it.

you know the radius of the circle is 1 with origin at (0,0).

when you draw it out in paper, you would have a right triangle in the 4th quadrant with 1/4 of the circle coming in the way.

The shortest distance would be the line perpendicular to the line Y=3/4X-3 and the circle.

so what I did was draw out a rectangle at the 4th quadrant (basically two 3-4-5 triangles) (0,0)(4,0)(-3,0)(4,-3)

Then I drew two diagonals with the length of 5 within the rectangle. The distance from the origin(0,0) to the midpoint of the rectangle is approximately 2.5.

the radius is 1, so you subtract 2.5 from 1 to get ~1.5 which is closest to A.

Like I said, my method was crude and I am not even sure if it right...maybe someone has a better explanation.
Join the discussion

by Claret » Mon Jun 29, 2009 7:54 pm
perpendicular Distance(shortest distance) of a point (x1,y1) from a line ax+by+c = 0 is given by
d = |(a*x1)+(b*y1)+c|/(a^2+b^2)^1/2

Let us calculate the distance of the line from the center of the circle (0,0)
a = -3/4
b=1
c=3
x1=0
y1=0

therefore d= 12/5

distance from a point on the circle would be 12/5-1 = 1.4

hope this helps!!
Last edited by Claret on Mon Jun 29, 2009 8:08 pm, edited 1 time in total.
Join the discussion

by Claret » Mon Jun 29, 2009 8:06 pm
ssmiles08 wrote:
the radius is 1, so you subtract 2.5 from 1 to get ~1.5 which is closest to A.
one thing to note here in this question is two answer choices are pretty close

A) 1.4 and B) (2) ^1/2 ~ 1.414

if we bring down the calculation to ~1.5 , either A or B could be the answer..
Join the discussion

by ssmiles08 » Tue Jun 30, 2009 7:52 am
Claret wrote:
ssmiles08 wrote:
the radius is 1, so you subtract 2.5 from 1 to get ~1.5 which is closest to A.
one thing to note here in this question is two answer choices are pretty close

A) 1.4 and B) (2) ^1/2 ~ 1.414

if we bring down the calculation to ~1.5 , either A or B could be the answer..
Yes you are right. Thank you.
Join the discussion

by boysangur » Tue Sep 07, 2010 9:45 pm
O_O ...this is a possible GMAT problem?

I don't even know where to start. What does "circle x^2 + y^2 = 1" even mean??
Join the discussion

by RCV » Tue Sep 07, 2010 11:06 pm
sanju, Ian, Kowinsky, please comment, found it very difficult for gmat
MBA admissions in leading universities of USA, Canada, UK, Australia, New Zealand, Singapore, Dubai.
Contact details: (+91)522-2200123 (+91)9369336687
Email: [email protected]
Located at Lucknow, India
Join the discussion

by sanjayviti » Wed Sep 08, 2010 3:19 am
The answer is indeed 1.4 but I think it needs to be solved differently.

I think everybody has figured out the vertices of traingle OAB that is O (0,0), A(0,-3) and B(4,0). Now draw a perpendicular line from O to AB (Sorry I have not drawn fig). Say it touches OB at D.

Now you can find out OD= OB*Sin <OAB

= 4*3/5
= 12/5

We know that a tangent on any point on the circle is perpendicular to the radius.

Therefore, the shortest distance from any point on circle to the line AB= OD- Radius of the circle

i.e. 12/5-1= 1.4
Join the discussion