A circle with center (1, 0) and radius 2 lies in the

If you draw a circle, you can solve it quickly.

We can see that for the 2nd quadrant, the (x,y) coordinate must be between \((0,\sqrt{3})\) and \((-1,0)\).

As we can clearly see that, if the radius is 2 and center is at (1,0); the extreme value of x for 2nd quadrant will be 1-2=-1, where y will be 0.

We know the extreme value for x; i.e. -1

Let's find out the extreme value for y; y will be maximum when x=0;

We can find y where x=0 by Pythagoras theorem;

Distance between origin and the center is 1.
Distance between center and the intersection of circle and y-axis is 2(radius).

Thus, distance between origin and extreme value of y in 2nd quadrant = \(\sqrt{2^2-1^2}=\sqrt{3}\approx 1.73\)

Thus we know the extreme(maximum) value for both x and y in the 2nd quadrant.

\(x<=-1 & y<=1.73\)

1. |y| > 1

Thus y can be 1.2 or 1.8.
1.2<1.73
1.8>1.73

(x,y) may or may not be in the 2nd quadrant
Not Sufficient.

2. |x| > 1

As we know that the maximum value of x in 2nd quadrant is 1. (x,y) can't be in the 2nd quadrant because |x|>1.

Sufficient.

Ans: "B"
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Please take a look at the image to get a visual clarification as well.

Thanks Fluke. Great explanation.

Interesting! Consider this we move the center to origin then both statements are insufficient. But not now.

Posted from my mobile device

Statement 1

Could be any quadrant actually

Statement 2

| x | > 1 means either x>1 or x<-1

Since circle has radius 2 and centered at (1,0) then x cannot be less than -1.

Therefore, x>1

So it CANNOT be on the second quadrant

Hence, answer is B only

Hope it helps!
Cheers!
J

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