A circle exists such that it touches both the y-axis and the

Question

A circle exists such that it touches both the y-axis and the line with equation x=y. What is the y-coordinate of the center of the circle ?

1. The x-coordinate of the center of the circle is 1.
2. The area of the circle is pi.

HowManyToGo · Answer

I vote for A. The stem says that the circle is either in the 1st or 3rd quadrant. 1) this says that the circle is in the 1st quadrant, and the radius is 1.There is only one way to place a circle of radius with 1 unit b/w the y-axis and the line x=y, so there can be only one value for Y.( can someone please let us know as to how to calculate the y coordinate here :?:

) 2)says that the radius is 1,don't know whether in the 1st or 3rd quadrant. HMTG> HMTG.

gmat2me2 · Answer

Agree with A....

It establishes that the circle is in quardant 1.

tkirk32 · Answer

orginally said D but realize that y=x could be first or third quadrant. DOH!

Agree with A.

Should wording of the question say "tangent to" instead of touches?

emilem · Answer

Agree with the fact there is a unique value of y in the 1st statement. However, the question asks what is the value of y? We can't find the specific value from given information. Lemme explain why.
The equation of circ is (x-a)^2 + (y-b)^2=R^2, where O(a,b) - center, M(x,y) - point on the circle, and R is radius.
Now let's plug-in values in 1st statement. For simplicity, assume that we choose tangent point on y-axis. Then, a=1, R=1, x=0. What follows, 1+(y-b)^2=1 or y=b. So, we still don't know the exact value of y.

Could you, guys, elaborate on this point?

sparky · Answer

a=1
(x1-a)^2+(y1-b)^2 = r^2
y1=x1
dy/dx=-(x1-a)/(y1-b) = 1
x2=0
(x2-a)^2+(y2-b)^2 = r^2
r=a

solve for a,b,r,x1,x2,y1,y2

tamg08 · Answer

IMO E,

I think that a circle with radius 1 and center (1,0) fits 1) and 2) and the question statement.

So wouldn't a circle with radius 1 and center (1, -.2), or (1, .5) also fit?

jallenmorris · Answer

I agree with E.

First, x=y....x could be -2 and y could be -2 or x = -3, y=-3...if both of those points are on the line made by x=y, it's still the same slope. The circle can be in Q1 or Q3 (top right and bottom left just in case I got the Q's numbered incorrectly).

With the second one, you know that it is  pi  distance from the y-axis, and you could figure out what options you have for coordinates with both statements, but nothing in the 2 statements will tell you which of the 2 possible quandrants the circle is in.

durgesh79 · Answer

the co-ordinate of the center of the circle = (1,k) 
we know radius of the circle is 1

this circle also touches line x-y=0

For a line with equation Ax + By + C = 0 and a point (r,s) the distance from the point to
the line is given by abs (Ar + Bs + C)/sqrt(A^2 + B^2)

we know this distance is radius = 1
 (1-k) / sqrt(2) = +/-1 
we can have two values of k
 k = 1+sqrt(2)  
and 
 k = 1-sqrt(2)

there can be two circles with the same radius,

statement 2 : area = pi --> radius 1 --> Not suff

Together not suff

Option E

greenoak · Answer

Another approach, without formulae (but requires drawing a picture): circles.gif Comments: Yellow lines – the locus of centres of the circles touching OY and y=x. So, we have two possibilities for the circle with the centre in (1, smth), and we have four possibilities for the circle with radius 1 – this means each St1 or St2 is insuff. For St 1+St 2, we still have two possibilities (as in the pic) => E.

A circle exists such that it touches both the y-axis and the

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