A circle in the coordinate plane passes through points (-3, -2) and (1

Question

A circle in the coordinate plane passes through points (-3, -2) and (1, 4). What is the smallest possible area of that circle?

A. 13π
B. 26π
C. 262√π
D. 52π
E. 64π

BrainLab · Accepted Answer

the given coordinates we can build a triangle. Let's say the chord is the diameter, so let's calculate the hypotenuse
4^2+6^2=52, area=Pi*r^2=13Pi Answer (A)

ENGRTOMBA2018 · Answer

Circle with minimum area will be when the 2 coordinates become the ends of the diameter. In all other cases, the coordinates will end up being the chord of the circle, thus giving you circles with larger areas (as dia = greatest chord in a circle).

Based on the given coordinates, the distance between the coordinates =  \sqrt{(-3-1)^2+(-2-4)^2}  =  \sqrt{52}

Thus, area = \frac{\pi*d^2}{4} = 13 \pi  . A is the correct answer.

peachfuzz · Answer

A circle in the coordinate plane passes through points (-3, -2) and (1, 4). What is the smallest possible area of that circle?

The smallest possible area for the circle will be one that has the distance between the 2 aforementioned points as the diameter.

The distance between the two points is sqrt(52). 
radius = sqrt(52)/2
Area= pi*(sqrt(52)/2)^2

Answer:
A. 13π

longfellow · Answer

Answer is 13pi.

The minimum area would be if these points are the endpoints of a diameter. using the distance formula or constructing a triangle and calculating the value of hypotenuse, we get the value of distance of the diameter and accordingly calculate the value of area.

Bunuel · Answer

VERITAS PREP OFFICIAL SOLUTION:

This problem requires you to leverage the definition that the longest chord in any circle is the diameter. Since the line connecting the two given points on the circle, (-3, -2) and (1, 4), is a chord, the smallest possible circle would occur if that chord were the diameter (the longest possible line).

Given that, taking the distance between (-3, -2) and (1, 4) will provide you with the diameter of that smallest possible circle. Since the horizontal difference (between x-coordinates) is 4 and the vertical difference (between y-coordinates) is 6, you can calculate the distance using Pythagorean Theorem:

4^2+6^2=c^2, so c^2=52 and  c=\sqrt{52}=2\sqrt{13} .

Remember that this distance is the diameter, so to find the area you'll want to cut it in half to find the radius. Therefore the radius is  \sqrt{13}  and  \pi r^2=\pi (\sqrt{13})^2=13\pi .

hatemnag · Answer

I selected A. but sorry Bunuel, I can not understand your point here. is there any drawing. I can not imagine. Sorry again.

ENGRTOMBA2018 · Answer

Let me try. When you draw series of circles through the points (-3,-2) and *(1,4), you will see that the line connecting these 2 points can be either a chord or the diameter (refer to the attached figure) 4-11-16 2-57-19 PM.jpg If the line is a chord, then the radius of this circle will be > the distance between the 2 points (= 2\sqrt {13} ) But, if the line is the diameter of the circle, then the distance = diameter of the circle. As area of a circle is dependent on the diameter of the circle, the smallest area of the circle will be when the line joining the 2 points above is the diameter of the circle. Thus, the circle with minimum area will have a diameter = 2\sqrt {13} and thus Area = \frac{\pi *(2\sqrt {13})^2}{4} = 13*\pi

hatemnag · Answer

Great analysis. thank you pretty much Engr2012.

Abhishek009 · Answer

Distance = Diameter =  \sqrt{(4+2)^2 + (1+3)^2}=>\sqrt{52}=2\sqrt{13}

Thus, Radius  = \sqrt{13}

Hence area of the circle  = π*\sqrt{13}*\sqrt{13} = 13π , Answer must be (A)

user34745792203 · Answer

Are these kind of circle exercises, as well as triangle exercises part of the GMAT Focus Edition?

user34745792203 · Answer

Are these kind of circle exercises, as well as triangle exercises part of the GMAT Focus Edition?

A circle in the coordinate plane passes through points (-3, -2) and (1

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GMAT Problem Solving (PS) Questions

A circle in the coordinate plane passes through points (-3, -2) and (1

Prep Toolkit

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My Rewards