abhi332 wrote:Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?

(1) a^2+b^2>16
(2) a=|b|+5

I believe the answer is (a) but I am not totally sure.

What you really want to know is, given (1) or (2), does the above equation definitely have a solution at x=0 (the y intercept)

For (1), let x=0. You have

x^2 + y^2 - 2yb + b^2 = 16
y^2 - 2yb = 16 - (a^2 + b^2)

since a^2 + b^2 = 16 from (1), we know that the right hand side of that equation has to be negative. In other words

y^2 - 2yb = some number less than zero
y^2 - 2yb < 0

y^2 < 2yb

either

y < 2b if y is positive
y > 2b if y is negative

There is a definite solution for these two possibilities.

To check for (2), observe the 2 situations

a = b+5 if b is positive
a = 5-b if b is negative

(b+5)^2 + (y-b)^2 = 16
(y-b)^2 = 16 - (b+5)^2

This obviously has a solution when b=-1 because that reduces to

(y+1)^2 = 0, which y =-1

But note that this has no solution when 16-(b+5)^2 < 0 because (y-b)^2 will always be a positive number. So this does NOT always yield the same answer. Thus (2) does not provide enough information.

abhi332 wrote:Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?

(1) a^2+b^2>16
(2) a=|b|+5

(x-a)^2 + (y-b)^2=16

the curve is a a circle with a radius of 4 having center at (a,b)

now for this to cut the y axis |a| <4

1) Doesn't help - Insufficient

2) a >5; sufficient the curve doesnt cut the y axis

B

ajith wrote:

abhi332 wrote:Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?

(1) a^2+b^2>16
(2) a=|b|+5

(x-a)^2 + (y-b)^2=16

the curve is a a circle with a radius of 4 having center at (a,b)

now for this to cut the y axis |a| <4

1) Doesn't help - Insufficient

2) a >5; sufficient the curve doesnt cut the y axis

B

That's a great way of looking at it. I didn't really see it that way.

ajith wrote:

abhi332 wrote:Does the curve (x-a)^2 + (y-b)^2=16 intersect the y-axis ?

(1) a^2+b^2>16
(2) a=|b|+5

(x-a)^2 + (y-b)^2=16

the curve is a a circle with a radius of 4 having center at (a,b)

now for this to cut the y axis |a| <4

1) Doesn't help - Insufficient

2) a >5; sufficient the curve doesnt cut the y axis

B

Ajith, I am not sure if it is a stupid question, but how did you find that the curve is a circle. Also, after identifying its a circle how did you solve the rest of the problem.

I will really appreciate if you can explain this in more detail. I am not very good in XY plane Q's.

Vineet

x*2 + y*2 = a*2 is the standard form of a circle.. (Read * as power of)


Cheers,
Siva
GMAT Aspirant

and (x-a)*2 + (y-a)*2 = a*2 is a circle with centre at (a,b)

In this qs within the scope of GMAT ?

kstv wrote:In this qs within the scope of GMAT ?

I do not think so

I can see 2 ways to solve it
1. Using geometry (involves equation of a circle etc etc)
2. Using Quadratic Equations (Involves finding out whether quadratic equation has real roots etc..)

Both of which are beyond the scope of GMAT.

ajith wrote:

kstv wrote:In this qs within the scope of GMAT ?

I do not think so

I can see 2 ways to solve it
1. Using geometry (involves equation of a circle etc etc)
2. Using Quadratic Equations (Involves finding out whether quadratic equation has real roots etc..)

Both of which are beyond the scope of GMAT.

While I agree this isn't a "standard" GMAT problem, solving quadratics is certainly within the scope of the GMAT. I don't think its easy but it certainly COULD be asked.

@ Ajith

acc to yr equations , u can assume that point a and point b will be less than 4 always. N this may be or may not is not the case here.


they can be any point. Let say a , b are 8, 10 then even radius is 4 then ofcourse do not cut Y axis but let say

point is 1 and 2 then circle will cut Y axis.



plz put OA

[spoiler]OA:B[/spoiler]

girish3131 wrote:@ Ajith

acc to yr equations , u can assume that point a and point b will be less than 4 always. N this may be or may not is not the case here.

plz put OA

When did I make that assumption?

I only said radius is 4 and the circle will cut y axis if the absolute value of a is less than 4. These are not assumptions, these are facts [There is a slight difference]

a,b can be anything

and 2) proves that a is greater than 5 - again no assumptions -

I am not getting the "assumption" you talk about here

Ajit have solved this question beautifully

another way is

curve will touch the y axis when x =0

therefore, a^2 + (y-b)^2 = 16

y^2 - 2yb + a^2 + b^2 -16 = 0

b^2 - 4ac >= 0, checking the roots

4b^2 - 4(1)(a^2 + b^2 -16) >=0
a^2 <=16

a^2-16<=0
(a+4)(a-4)<=0

a<= -4 or a<= 4

|a| <= 4


which is same what Ajit has found quickly

What I understand from this question is that we need to know the distance between the center point of the circle and the y - axis i.e the y - intercept of the point.

(2) a=|b|+5

from this we do not care about the value for 'a' which is the x-intercept.

So could any of the experts please guide whether i m correct or wrong.

I am getting the value b as + or - 5. Hence the correct answer is B.

Could any of you help...