GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 18 Aug 2019, 12:21

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Line K is tangent to the circle with center (0, 0) at (-3, y) such tha

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 57030
Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:00
4
20
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

55% (01:44) correct 45% (02:04) wrong based on 413 sessions

HideShow timer Statistics

Line K is tangent to the circle with center (0, 0) at (-3, y) such that its y-intercept is positive. If the radius of the circle is 5, what is the slope of Line K?


A. \(-\frac{4}{3}\)

B. \(-\frac{3}{4}\)

C. \(\frac{3}{4}\)

D. \(\frac{1}{2}\)

E. Cannot be determined

 

This question was provided by Crack Verbal
for the Game of Timers Competition

 


_________________
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 57030
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 19 Jul 2019, 01:58
1
Bunuel wrote:
Line K is tangent to the circle with center (0, 0) at (-3, y) such that its y-intercept is positive. If the radius of the circle is 5, what is the slope of Line K?


A. \(-\frac{4}{3}\)

B. \(-\frac{3}{4}\)

C. \(\frac{3}{4}\)

D. \(\frac{1}{2}\)

E. Cannot be determined

 

This question was provided by Crack Verbal
for the Game of Timers Competition

 



OFFICIAL EXPLANATION: FROM CRACK VERBAL:



Since the circle is centered at (0, 0), has a radius of 5 and is tangent at (-3, y), using the distance formula we can find the value of y.

5^2 = (-3 – 0)^2 + (y – 0)^2 ---> 25 = 9 + y^2 --> y^2 = 16 --> y = 4

Now to find the slope of line K ideally we will require two points on the line, but we know that the radius and the tangent in a circle are always going to be perpendicular to each other. If two lines are perpendicular, the product of the slopes will always be equal to -1.

Here line K and the radius are perpendicular to each other and we have two points on the radius i.e. (0, 0) and (-3, 4). The slope of the radius will be -4/3. Since line K is perpendicular to the radius, the slope of line K will be ¾.
_________________
Most Helpful Community Reply
Manager
Manager
User avatar
G
Joined: 22 May 2015
Posts: 123
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:18
8
1
Slope of two perpendicular lines is -1.

Distance between the points is 5 , hence solving for y

(-3) ^ 2 + y^2 = 5^2

y = 4

Hence the equation of line is

Since y=mx+c , using the points 0,0 and -3,4 we get m = -4/3. Hence slope of line will be 3/4. Option C
Attachments

gg.jpg
gg.jpg [ 3.81 MiB | Viewed 3854 times ]


_________________
Consistency is the Key
General Discussion
Manager
Manager
avatar
G
Joined: 27 May 2010
Posts: 200
Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:11
5
Radius is given to be 5.

Tangent would meet the circle at a point such that it is perpendicular to the radius drawn from the centre to the point of intersection.

Given that the tangent meets the circle at (-3,y).

Using distance formula (0-(-3))^2 + (0-y)^2 = 5^2
y^2 = 16
y = +/- 4

Slope of radius line= (y2-y1)/(x2-x1) = (0+/-4)/(0-3)
=+/- 4/3

Slope of tangent*Sloe of radius line = -1
Slope of tangent = 3/4 or -3/4

So, it cannot be determined
Option E.
_________________
Please give Kudos if you like the post
Manager
Manager
avatar
S
Joined: 25 Aug 2016
Posts: 54
Location: India
WE: Information Technology (Computer Software)
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:17
1
2
(x−a)^2+(y−b)^2=r^2

(-3-0)^2+(y-0)^2=5^2
9+y^2=25
y^2=25-9=16
y=+4/-4
As Y-intercept in positive,y=4
slope=-4/3
Manager
Manager
avatar
G
Joined: 31 May 2018
Posts: 183
Location: United States
Concentration: Finance, Marketing
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:20
1
1
since centre of circle (0,0) radius=5
equation of circle \((x-a)^2+(y-b)^2\)=\(r^2\) (where a,b centre of circle and r radius)
\(x^2+y^2=25\)
line k is tangent at (-3,y) so this point lies on circle
now putting values of x and y in equation
9+\(y^2\)=25
\(y^2\)=16
y has two values +4 and -4
we will take y=4 ( since given that line k has positive y intersept)
now we can find the slope
points are (-3,4)
slope=tan θ (θ is angle)=\(\frac{y}{x}\)=\(\frac{4}{-3}\)
A is the answer
Intern
Intern
avatar
B
Joined: 04 Feb 2019
Posts: 5
Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post Updated on: 01 Jul 2019, 08:24
1
1
Given - Point of tangent to the circle with center at O(0,0) is A(-3,y) and the Radius is 5.

So OA^2 = (0 - (-3))^2 + (0 - y)^2 = 5^2
Gives y=4 (Not -4 since y intercept is positive)

Let slope of OA be m1 = (4-0)/(-3-0) = -4/3

Slope of tangent is m2 = -1/m1 (Property of slopes of perpendicular lines) = 3/4

Answer C

Originally posted by taransaurav on 01 Jul 2019, 08:23.
Last edited by taransaurav on 01 Jul 2019, 08:24, edited 1 time in total.
Manager
Manager
avatar
B
Joined: 10 Jun 2019
Posts: 51
CAT Tests
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:23
2
We need to establish 2 baselines here : If the line K is tangent to the circle,It is perpendicular to the radius at the point of tangency. If the y intercept of line K is positive it means Line K exists in Quadrant I and II and maybe Quadrant III. To find y we find the distance between the centre of the circle and the point (-3,y). Which gives us y^2=16. y could be 4/-4 but because line K is tangent to the circle in Quadrant I, y is 4. Now we need to find the slope of the radius which will be the negative reciprocal of line K. Finding the slope of the radius/line with points (0,0) and (-3,4) gives -4/3.The negative reciprocal of this will be 3/4.Thus C
Director
Director
avatar
G
Joined: 22 Nov 2018
Posts: 532
Location: India
GMAT 1: 640 Q45 V35
GMAT 2: 660 Q48 V33
Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post Updated on: 01 Jul 2019, 08:24
Since the line is tangent to the circle with center at (0,0) the distance between the two points is equal to radius; Squareroot[(y-0)^2+(-3-0)^2]=5; squaring on both sides gives y^2+9=25; y^2=16; y=4. Slope of line -4/3 (using (y2-y1)/(x2-x1) ) IMO A
_________________
Give +1 kudos if this answer helps..!!

Originally posted by Arvind42 on 01 Jul 2019, 08:24.
Last edited by Arvind42 on 01 Jul 2019, 08:24, edited 1 time in total.
Verbal Forum Moderator
User avatar
V
Status: Greatness begins beyond your comfort zone
Joined: 08 Dec 2013
Posts: 2386
Location: India
Concentration: General Management, Strategy
Schools: Kelley '20, ISB '19
GPA: 3.2
WE: Information Technology (Consulting)
GMAT ToolKit User Reviews Badge CAT Tests
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:24
3
1
Line K is tangent to the circle with center (0, 0) at (-3, y) such that its y-intercept is positive. If the radius of the circle is 5, what is the slope of Line K?


A. \(-\frac{4}{3}\)

B. \(-\frac{3}{4}\)

C. \(\frac{3}{4}\)

D. \(\frac{1}{2}\)

E. Cannot be determined

Line K is tangent to the circle with center (0, 0) at (-3, y) such that its y-intercept is positive

Since y-intercept is positive--> line rises as it moves towards the right
(-3,y) lies on the circumference of the circle with radius 5
y= 4
x^2+y^2= r^2

Line K is perpendicular to line L(that joins point (-3,4) and center of circle(0,0)
Slope of line L = -4/3

Slope of line K = 3/4 (Product of slope of perpendicular lines is -1)

Answer C
_________________
When everything seems to be going against you, remember that the airplane takes off against the wind, not with it. - Henry Ford
The Moment You Think About Giving Up, Think Of The Reason Why You Held On So Long
Intern
Intern
avatar
B
Joined: 09 Mar 2019
Posts: 11
CAT Tests
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:25
1
circle has radius of 5 unit.
Y Intercept is positive so we take that point is in the second quadrant. So this point where tangent touches point (-3,Y)
5^2 = (-3-0)^2+(Y-0)^2
Y = 4
so slope of line joining point (-3, 4) and (0,0) = difference of Y points/difference of X points = (Y1 - Y2)/(X1-X2) = (4 - 0)/(-3- 0)=-4/3
and line tangent to this point will be perpendicular to this line so slope of the line K will be inverse and opposite in sign to the tangent joinining line (-3,4 ) and (0,0) and that will be equal to = 3/4
so IMO ans is C
Manager
Manager
avatar
S
Joined: 24 Jan 2019
Posts: 103
Location: India
Concentration: Strategy, Finance
GPA: 4
Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:27
1
Radius is 5 and touching point is (-3,y).
Here y can be +4 or -4.
Since y-intercept is positive, y will be 4.
So, line joining the touching point and center will have slop of -4/3. Tanget will be perpendicular to this line with slop 3/4.

Answer : C

Posted from my mobile device
General GMAT Forum Moderator
User avatar
V
Joined: 15 Jan 2018
Posts: 508
Concentration: General Management, Finance
GMAT 1: 720 Q50 V37
Premium Member Reviews Badge
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:34
5
Given that,

Radius of the circle = 5
Tangent K meets the circle with center (0, 0) at (-3, y)
According to Tangent Theorem, the radius from the center of the circle to the point of tangency is perpendicular to the tangent line.

Therefore, we can solve for y, using the Pythagoras theorem.
\(-3^{2}\)+ \(y^{2}\) = \(5^{2}\)
\(y^{2}\) = 16
y = 4
Note: Since the question specifives the y intercept is positive, we can ignore the negative value of y, i.e. y = -4

Now using the slope equation of a straight line, y = mx + c, where m is slope and c is the y-intercept, we can find the equation of the radius
The slope of the radius is \(\frac{-4}{3}\)

The product of the slopes of two perpendicular lines is -1
Therefore, (slope of radius) x (slope of line k) = -1

Slope of line k = \(\frac{3}{4}\)

The correct answer is option C. \(\frac{3}{4}\)
_________________
In case you find my post helpful, please provide me with kudos. Thank you!

New to GMAT Club or overwhelmed with so many resources? Follow the GMAT Club Study Plan!
Not happy with your GMAT score? Retaking GMAT Strategies!
Manager
Manager
avatar
P
Joined: 01 Aug 2017
Posts: 223
Location: India
Concentration: General Management, Leadership
GMAT 1: 500 Q47 V15
GPA: 3.4
WE: Information Technology (Computer Software)
Reviews Badge CAT Tests
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:37
Radius is 5. Which means cordinate is (-3,4).

Using slope formula (y2 -y1)/(x2-x1)

(4-0)/(-3-0) = -4/3

Ans A

Posted from my mobile device
_________________
If it helps you please press Kudos!

Thank You
Sudhanshu
Manager
Manager
avatar
P
Joined: 01 Aug 2017
Posts: 223
Location: India
Concentration: General Management, Leadership
GMAT 1: 500 Q47 V15
GPA: 3.4
WE: Information Technology (Computer Software)
Reviews Badge CAT Tests
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:38
Radius is 5. Which means cordinate is (-3,4).

Using slope formula (y2 -y1)/(x2-x1)

(4-0)/(-3-0) = -4/3

Ans A

Posted from my mobile device
_________________
If it helps you please press Kudos!

Thank You
Sudhanshu
Senior Manager
Senior Manager
avatar
P
Joined: 05 Jul 2018
Posts: 340
Location: India
Concentration: General Management, Technology
GMAT 1: 600 Q47 V26
GRE 1: Q162 V149
GPA: 3.6
WE: Information Technology (Consulting)
Reviews Badge CAT Tests
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:39
3
The correct answer is C (3/4)

How?

Well, the tangent touches the circle at one point and is perpendicular to the radius. That point is given as (-3,y) and we also know that radius is 5 units.

Now for the radius to be 5 units . Let us consider the radius to be a line with 2 points A, B. A is the origin (0,0) and B is (-3,y). And distance between A and B is 5

Therefore,\(\sqrt{(x2-x1)^2 +(y2-y1)^2} =5\)
\(3^2 +y^2=5^2\)
\(y^2=16\)
\(y=4\)

Slope of line AB = \(\frac{(y2-y1)}{(x2-x1)}\) = \(\frac{(4)}{(-3)}\) = \(\frac{-4}{3}\)

Radius and tangent are perpendiculars

Product of slope of 2 perpendicular lines = m1*m2=-1

\(\frac{-4}{3} * (Slope of Tangent) =-1\)
Hence. The slope of Tangent =\(\frac{3}{4}\)
_________________
Manager
Manager
avatar
S
Joined: 08 Jan 2018
Posts: 98
Location: India
GPA: 4
WE: Information Technology (Computer Software)
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:41
2
Hi,

As given y>0, and tangent touches at (-3,y), and the radius = 5.
We can use the distance formula between points to find y.
\(\sqrt{{(-3-0)^2} + {(y-0)^2}}\) = \(\sqrt{{5^2}}\)

This will give us y= +4 or -4 , since we know y> 0, so y= +4.

Now, the point is (-3,4). We can use the line equation between points:

(y-0) = \(\frac{(4-0)}{(-3-0)}\) * (x-0)
y = (-4/3) x

As per the property of circle, the tangent should touch at 90-degree angle, so the slope formula is => m1 * m2 = -1
using the formula we get :
(-4/3) * m2 = -1
m2 = (3/4)

Ans = 3/4

Please hit Kudos if you like the solution.
Intern
Intern
avatar
B
Joined: 21 Feb 2019
Posts: 7
Location: India
Concentration: Operations, General Management
Schools: ISB '21
WE: Supply Chain Management (Manufacturing)
CAT Tests
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:47
Since radius is given we can find Y
y= 4 or y =-4 but given Y intercept as positive so Y = 4

Slope can be found by (y-0)/(-3-0)= -4/3
Intern
Intern
avatar
S
Joined: 27 Feb 2017
Posts: 43
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:50
1
This is a good question. I spent a bit too much time (2 minute and 46 seconds) on it as I had hoped to solve the question directly.

The best approach is: (1) Figure out that the (-3, y) is actually (-3, 4)
(2) So, the slope for the line between (-3,4) and (0,0) is -4/3
(3) The answer is -1 / (-4/3) = 3/4.

So, the answer is C.
Intern
Intern
User avatar
S
Joined: 24 Mar 2018
Posts: 48
Location: India
Concentration: Operations, Strategy
Schools: ISB '21
WE: Project Management (Energy and Utilities)
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha  [#permalink]

Show Tags

New post 01 Jul 2019, 08:50
2
Line K is tangent to the circle at the point (-3,y). Let’s call this point A.

The centre of the circle is C (0,0).

Let’s drop a perpendicular from A to the x-axis. The co-ordinate of the same is (-3,0). We’ll call this point B.

Joining the three points, we get a right angled triangle right angled at B. The sides are: AB = y, BC = 3, AC = 5 (radius)

Using Pythagoras theorem, we get y = 4. Hence, the point of tangency A is (-3,4).

The slope of the line AC is: (0-4)/(0+3) = -4/3

As line K is perpendicular to AC (property of tangency), the slope of K is the negative reciprocal, ie. 3/4.

Posted from my mobile device
GMAT Club Bot
Re: Line K is tangent to the circle with center (0, 0) at (-3, y) such tha   [#permalink] 01 Jul 2019, 08:50

Go to page    1   2   3   4    Next  [ 77 posts ] 

Display posts from previous: Sort by

Line K is tangent to the circle with center (0, 0) at (-3, y) such tha

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne