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# What is the equation of the circle C in which an equilateral triangle

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Director
Joined: 03 Jun 2019
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What is the equation of the circle C in which an equilateral triangle  [#permalink]

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09 Jul 2019, 01:15
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55% (hard)

Question Stats:

66% (02:27) correct 34% (02:47) wrong based on 56 sessions

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What is the equation of the circle C in which an equilateral triangle is inscribed which has lines L1 x =0, line L2 $$\sqrt{3}y+x=\sqrt{3}$$ and Line L3 $$\sqrt{3}y-x+\sqrt{3}=0$$ as its 3 sides?

$$A. (x-\sqrt{3}/2)^2 + y^2 = \frac{4}{3}$$
$$B. (x-\sqrt{3}/3)^2+y^2=\frac{4}{3}$$
$$C. x^2 +y^2 = 4$$
$$D. (x-\sqrt{3})^2 + y^2 = 4$$
$$E. (x-\sqrt{3}/4)^2+y^2=\frac{4}{3}$$

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Re: What is the equation of the circle C in which an equilateral triangle  [#permalink]

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09 Jul 2019, 06:15
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Kinshook wrote:
What is the equation of the circle C in which an equilateral triangle is inscribed which has lines L1 x =0, line L2 $$\sqrt{3}y+x=\sqrt{3}$$ and Line L3 $$\sqrt{3}y-x+\sqrt{3}=0$$ as its 3 sides?

$$A. (x-\sqrt{3}/2)^2 + y^2 = \frac{4}{3}$$
$$B. (x-\sqrt{3}/3)^2+y^2=\frac{4}{3}$$
$$C. x^2 +y^2 = 4$$
$$D. (x-\sqrt{3})^2 + y^2 = 4$$
$$E. (x-\sqrt{3}/4)^2+y^2=\frac{4}{3}$$

Line x = 0 is the y axis.
The two lines will intersect the y axis when x = 0
$$\sqrt{3}y+ 0 =\sqrt{3}$$
y = 1
$$\sqrt{3}y-0+\sqrt{3}=0$$
y = -1

So the circle will pass through (0, 1) and (0, -1). Only options (B) and (D) will satisfy these points.
The side of the equilateral triangle will be 2. Now note that the radius of the circle cannot be 2 since the side of the equilateral triangle must be greater than the radius.
So option (B) must be correct.
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Joined: 26 Apr 2019
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Re: What is the equation of the circle C in which an equilateral triangle  [#permalink]

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09 Jul 2019, 04:47
imo B . i just found point of intersection of all 3 line and put into options to check if point lie on the circle . only b satisfies all 3 point of intersection
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What is the equation of the circle C in which an equilateral triangle  [#permalink]

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09 Jul 2019, 05:38
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Given, L1 x =0, line L2 3√y+x=3√3y+x=3 and Line L3 3√y−x+3√=03y−x+3=0
So, we can get the vertices value by the point of intersection of these lines.
Let triangle is ABC, so for getting A, B, C we should equate the lines.
=> L1 =L2 , L2 = L3, L3 = L1
So, upon solving we can get A (0,1) B (0,-1) and C($$\sqrt{3}$$,0)

A triangle is inscribed in the circle, then the distance from the center of the circle to all the vertices will be equal.

Let the Center of Circle is O(a,b).
Then using the distance formula:
AO = OB , OB = OC , OC = OA
Upon solving these equations we get a = ($$\sqrt{3}$$/3) and b= 0
For radius: The distance from the center to the Vertices will be the radius, as it`s equal.
So, radius = OA = OB = OC
upon solving any distance:

Thus, using circle formula: $$(x-a)^2$$ + $$(y-b)^2$$ = $$(4/3)^2$$

Thus putting the values we get the answer as B.

Please hit the Kudos if you like the solution.
Director
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What is the equation of the circle C in which an equilateral triangle  [#permalink]

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09 Jul 2019, 12:27
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Intersection point of $$L_1$$ and $$L_2$$ is (0, 1) and that of $$L_1$$ and $$L_3$$ is (0, -1)

Center of the circle lies on perpendicular bisector of line $$L_1$$
Midpoint of Line $$L_1$$ is (0, 0).
In center of the triangle= $$\frac{1}{2\sqrt{3}}$$*2= $$\frac{1}{\sqrt{3}}$$

Co-ordinates of center= ($$\frac{3}{\sqrt{3}}$$, 0)

Radius of circle= $$\frac{1}{\sqrt{3}}$$*2=$$\frac{2}{\sqrt{3}}$$

Kinshook wrote:
What is the equation of the circle C in which an equilateral triangle is inscribed which has lines L1 x =0, line L2 $$\sqrt{3}y+x=\sqrt{3}$$ and Line L3 $$\sqrt{3}y-x+\sqrt{3}=0$$ as its 3 sides?

$$A. (x-\sqrt{3}/2)^2 + y^2 = \frac{4}{3}$$
$$B. (x-\sqrt{3}/3)^2+y^2=\frac{4}{3}$$
$$C. x^2 +y^2 = 4$$
$$D. (x-\sqrt{3})^2 + y^2 = 4$$
$$E. (x-\sqrt{3}/4)^2+y^2=\frac{4}{3}$$
Manager
Joined: 29 Dec 2018
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What is the equation of the circle C in which an equilateral triangle  [#permalink]

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16 Jul 2019, 19:55
[quote
So the circle will pass through (0, 1) and (0, -1).Only options (B) and (D) will satisfy these points.
The side of the equilateral triangle will be 2. Now note that the radius of the circle cannot be 2 since the side of the equilateral triangle must be greater than the radius.
So option (B) must be correct.[/quote]

VeritasKarishma can you please explain this further? how did option B & D satisfy 0,1 and 0, -1
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Re: What is the equation of the circle C in which an equilateral triangle  [#permalink]

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17 Jul 2019, 16:27
Kinshook wrote:
What is the equation of the circle C in which an equilateral triangle is inscribed which has lines L1 x =0, line L2 $$\sqrt{3}y+x=\sqrt{3}$$ and Line L3 $$\sqrt{3}y-x+\sqrt{3}=0$$ as its 3 sides?

$$A. (x-\sqrt{3}/2)^2 + y^2 = \frac{4}{3}$$
$$B. (x-\sqrt{3}/3)^2+y^2=\frac{4}{3}$$
$$C. x^2 +y^2 = 4$$
$$D. (x-\sqrt{3})^2 + y^2 = 4$$
$$E. (x-\sqrt{3}/4)^2+y^2=\frac{4}{3}$$

Line x = 0 is the y axis.
The two lines will intersect the y axis when x = 0
$$\sqrt{3}y+ 0 =\sqrt{3}$$
y = 1
$$\sqrt{3}y-0+\sqrt{3}=0$$
y = -1

So the circle will pass through (0, 1) and (0, -1). Only options (B) and (D) will satisfy these points.
The side of the equilateral triangle will be 2. Now note that the radius of the circle cannot be 2 since the side of the equilateral triangle must be greater than the radius.
So option (B) must be correct.

Can you explain how did you calculate the diameter in D?
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Re: What is the equation of the circle C in which an equilateral triangle  [#permalink]

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17 Jul 2019, 22:19
nick1816 wrote:
Intersection point of $$L_1$$ and $$L_2$$ is (0, 1) and that of $$L_1$$ and $$L_3$$ is (0, -1)

Center of the circle lies on perpendicular bisector of line $$L_1$$
Midpoint of Line $$L_1$$ is (0, 0).
In center of the triangle= $$\frac{1}{2\sqrt{3}}$$*2= $$\frac{1}{\sqrt{3}}$$

Co-ordinates of center= ($$\frac{3}{\sqrt{3}}$$, 0)

Radius of circle= $$\frac{1}{\sqrt{3}}$$*2=$$\frac{2}{\sqrt{3}}$$

Kinshook wrote:
What is the equation of the circle C in which an equilateral triangle is inscribed which has lines L1 x =0, line L2 $$\sqrt{3}y+x=\sqrt{3}$$ and Line L3 $$\sqrt{3}y-x+\sqrt{3}=0$$ as its 3 sides?

$$A. (x-\sqrt{3}/2)^2 + y^2 = \frac{4}{3}$$
$$B. (x-\sqrt{3}/3)^2+y^2=\frac{4}{3}$$
$$C. x^2 +y^2 = 4$$
$$D. (x-\sqrt{3})^2 + y^2 = 4$$
$$E. (x-\sqrt{3}/4)^2+y^2=\frac{4}{3}$$

_________________
"Success is not final; failure is not fatal: It is the courage to continue that counts."

Please provide kudos if you like my post. Kudos encourage active discussions.

My GMAT Resources: -

Efficient Learning

Tele: +91-11-40396815
Mobile : +91-9910661622
E-mail : kinshook.chaturvedi@gmail.com
Re: What is the equation of the circle C in which an equilateral triangle   [#permalink] 17 Jul 2019, 22:19
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