On the coordinate plane , points P and Q are defined by the

An alternative way to approach this problem would be to calculate the distance between the center point and the other two points mentioned in the problem. The distance between center point and other two given points should be the same (will be radius) .
Distance between two points can be calculated as square root of ((x2-x1)^2+(y2-y1)^2))

Option -1: Distance 1 is √(6.25+1 )=√7.25 , Distance 2 is √(2.25+4)=√6.25.Hence Incorrect.
Option-2: Distance1 is √(9+25 )=√34 , Distance 2 is √(1+64)=√65.Hence Incorrect.
Option-3: Distance1 is √(1+0 )=√1 , Distance 2 is √(9+9)=√18.Hence Incorrect.
Option-2: Distance1 is √(4+2.25)=√6.25 , Distance 2 is √(4+2.25)=√6.25. Both are equal and hence Correct.
Option-2: Distance1 is √(9+4 )=√13 , Distance 2 is √(1+1)=√2. Hence Incorrect.

really nice works everyone....................

I m sorry to ask this question.. I am new to GMAT, you may find this question silly but still I cant able to get this.
We have a point on the circle(x,y)=(-1,0) and we have radius r=5/2.
Equation of circle is : (x-a)^2+(y-b)^2 = r^2. By simply substituting above (x,y) and r we can find (a,b) which is the center.
Y WE ARE GETTING 2 EQUATIONS(as done by maaadhu) SOLVING IT TO GET A FINAL EQUATION AND THEN SUBSTITUTING ANSWER OPTIONS AND SEEING?

The approach you are talking about allows to get the answer without actually solving the equations.

BTW, how can you solve (-1-a)^2+(0-b)^2 = (5/2)^2 for a and b?

Yes brother you can back solve it from the last equation you got...
but evaluating the reals roots of the equation will be cumbersome.......

Indeed you can solve any gmat problem without applying some formulas ,And just with few basic understandings.

Thanks Bunuel and Asif for your replies. I got the gap where i fell.

Area of the circle
\(pi*r^2 = \frac{25}{4}pi\)
=> \(r=\frac{5}{2}\)

Distance between the given points is
\(\sqrt{(3-(-1))^2 + (3-0)^2}\)

\(=\sqrt{16+9}\)

\(=5\)

Hence the given points are end points of diameter & mid point of these two points is the center of the circle.

mid-point is
\((\frac{{3+(-1)}}{2},\frac{{3+0}}{2})\)

\(=(1,1.5)\)

Although it took me 3 mins to solve this question using all those equations, later I thought this question can be solved easily using options.

One property to keep in mind - A line passing through the centre of the circle bisects the chord (or passes from the mid point of the chord).

Now mid point of chord here is (-1+3)/2, (3+0)/2 i.e. (1,1.5) now luckily we have this in our Ans. choice. so definitely this is the ans. It also indictaes that PQ is the diameter of the circle.

There can be a case when PQ is not a diameter but in that case also the y-coordinate will remain same as it is the midpoint of the chord and we are moving up in the st. line to locate the centre of the circle.

If ans choices are all distinct (y cordinates) ONLY CHECK FOR Y CORDINATE and mark the ans.

Hi nailgmat2015,

Please note that every line passing through centre of the circle does not bisect the chord. Only the line which passes through the centre and is perpendicular to the chord bisects the chord. For example refer the diagram below. In this diagram there are various lines which pass through the centre and intersect the chord, but only OD will bisect the chord as it is perpendicular to the chord.



For this question, you got the right answer as you assumed PQ to be the diameter which is actually the case. A better method for this question would have been:

Step-I
Find the radius of the circle from the given area of the circle which gives us radius = 2.5 and hence diameter = 5

Step-II
In this case, we can observe that the length of the chord PQ is 5( by using the distance formula). Since the length of chord PQ is equal to the length of the diameter that would mean that PQ is the diameter of the circle.
Since PQ is the diameter of the circle, the midpoint of PQ would be the centre of the circle which gives us the coordinates of centre as (1, 1.5)

Hope its clear

Regards
Harsh

Hi harsh,

Thanks for such a wonderful explanation. I guess I need to brush up my basics once.

Thanks,
R

In the first explanation, after finding AC=5, how did we conclude that its a diameter? I am missing something here.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.