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Senior Manager  Joined: 06 Aug 2011
Posts: 329
Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)  [#permalink]

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Thanks again Bunuel _________________
Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !
Intern  Joined: 05 Feb 2014
Posts: 42
Re: Cord Geometry  [#permalink]

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lonewolf wrote:
Good method. Simple and quick.

Just for refreshing memories... If two chords of a circle form a right angle degree (for example: AB, BC), then the chord AC must be the diameter of the circle.

Is the above statement always true?

but how do we get to know that the right angle is formed ?
Math Expert V
Joined: 02 Sep 2009
Posts: 55267
Re: Cord Geometry  [#permalink]

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gauravsoni wrote:
lonewolf wrote:
Good method. Simple and quick.

Just for refreshing memories... If two chords of a circle form a right angle degree (for example: AB, BC), then the chord AC must be the diameter of the circle.

Is the above statement always true?

but how do we get to know that the right angle is formed ?

Check here: if-a-circle-passes-through-points-1-2-2-5-and-42105.html#p672194
_________________
Intern  Joined: 05 Feb 2014
Posts: 42
Re: Cord Geometry  [#permalink]

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Bunuel wrote:
gauravsoni wrote:
lonewolf wrote:
Good method. Simple and quick.

Just for refreshing memories... If two chords of a circle form a right angle degree (for example: AB, BC), then the chord AC must be the diameter of the circle.

Is the above statement always true?

but how do we get to know that the right angle is formed ?

Check here: if-a-circle-passes-through-points-1-2-2-5-and-42105.html#p672194

I understand that if a right triangle is formed then the side will be the diameter. But in the question how do we come to know that a right angle triangle is being formed ?
Math Expert V
Joined: 02 Sep 2009
Posts: 55267
Re: Cord Geometry  [#permalink]

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1
gauravsoni wrote:
Bunuel wrote:
gauravsoni wrote:
but how do we get to know that the right angle is formed ?

Check here: if-a-circle-passes-through-points-1-2-2-5-and-42105.html#p672194

I understand that if a right triangle is formed then the side will be the diameter. But in the question how do we come to know that a right angle triangle is being formed ?

I guess you did not read that post to the end:

The slope of line segment: A(1,2) and B(2,5) is 3 AND the slope of line segment: B(2,5) and C(5,4) is -1/3, the slopes are negative reciprocals hence these line segments are perpendicular to each other. We have right triangle ABC, AC=hypotenuse=Diameter.
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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)  [#permalink]

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Hi Friends,

This Q can b solved under a minute, as the question asks for the diameter, we know that the value must be 2*radius i.e. and among the options mentioned only sqrt(20) can be identified as 2*sqrt(5). Hence option B

Hopes it makes sense.
Manager  B
Joined: 28 Apr 2016
Posts: 87
Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)  [#permalink]

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Since the formula is (x-a)^2 + (y-b)^2 = r^2. How did you take (a-1)^2 + (b-2)^2 = r^2?

Fig wrote:
(B) for me The equation of circle C in the XY plan is :
(x-a)^2 + (y-b)^2 = r^2

Where :
o (a,b) is the center of the circle
o r is the radius of the circle

Knowing that, u can plug the coordonates of each points to get a, b and r.

(1, 2) on C implies:
(a-1)^2 + (b-2)^2 = r^2
<=> a^2 -2*a + 1 + b^2 - 4*b + 4 = r^2 (1)

(2, 5) on C implies:
(a-2)^2 + (b-5)^2 = r^2
<=> a^2 -4*a + 4 + b^2 - 10*b + 25 = r^2 (2)

(5, 4) on C implies:
(a-5)^2 + (b-4)^2 = r^2
<=> a^2 -10*a + 25 + b^2 -8*b + 16 = r^2 (3)

(1) - (2)
<=> 2*a - 3 + 6*b -21 =0
<=> 2*a + 6*b = 24
<=> a + 3*b = 12 (4)

(1) - (3)
<=> 8*a -24 + 4*b - 12 = 0
<=> 2*a + b = 9 (5)

(5) -2*(4)
<=> -5*b = -15
<=> b = 3

From (1), we find a :
a = 12 - 3*b = 12 - 3*3 = 3

Still from (1):
(a-1)^2 + (b-2)^2 = r^2
<=> (2)^2 + (1)^2 = r^2
<=> r = sqrt(5)

Thus,
D = 2*sqrt(5) = sqrt(20)
Manager  Joined: 05 Sep 2014
Posts: 73
Schools: IIMB
Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)  [#permalink]

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Bunuel wrote:
gottabwise wrote:
lonewolf: I'd like to know the answer as well. Great question.

I got B. Draw a coord plane, plotted pts, draw right triangle. Used distance formula for each side. Thought about the radius formula for circumscribed circles. Nixed idea. Noticed isoceles right triangle and realized sqrt 20 was diameter b/c it's the hypotenuse. Ended there. Glad I tried the problem.

A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.
Attachment:
Math_Tri_inscribed.png

So: If two chords of a circle form a right angle degree (for example: AB, BC), then the chord AC must be the diameter of the circle. TRUE.

As for the question: the slope of line segment: A(1,2) and B(2,5) is 3 AND the slope of line segment: B(2,5) and C(5,4) is -1/3, the slopes are negative reciprocals hence these line segments are perpendicular to each other. We have right triangle ABC, AC=hypotenuse=Diameter.

Also one tip: any three points, which are not collinear, define the unique circle on XY-plane. For more please see the Triangles and Circles chapters of Math Book in my signature.

Hope it's clear.

Hi Bunnuel,

Thanks a lot for your explanation, really helped me. However just wanted to understand how do we calculate the diameter if its not the right angle triangle in a circle. Apologies if this sounds too basic. I made a graph of the same, unable to analyze what could be diameter in such question.

Thanks in advance for your kind help.

Regards
Megha
Senior Manager  G
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Posts: 288
Daboo: Sonu
GMAT 1: 590 Q49 V20 GMAT 2: 730 Q50 V38 Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)  [#permalink]

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bz9 wrote:
If a circle passes through points $$(1, 2)$$, $$(2, 5)$$, and $$(5, 4)$$, what is the diameter of the circle?

A. $$\sqrt{18}$$
B. $$\sqrt{20}$$
C. $$\sqrt{22}$$
D. $$\sqrt{26}$$
E. $$\sqrt{30}$$

Let us consider the origin is at (x,y)
so from distance formula
(x-1)^2+ (y-2)^2 =r^2 ..................................1
(x-2)^2+(y-5)^2= r^2 ..................................2
(x-5)^2+(x-4)^2 =r^2 ..................................3
on solving 1 and 2 we get x+3y =12 ..................4
on solving 1 and 3 we get 2x+ 6y=9 ..................5
now solve 4 and 5
(x,y)= (3,3) and is the center
Sqrt((3-1)^2+(3-2)^2)= sqrt( 4+ 1)= sqrt(5)= radius
Diamiter = 2*radius = 2* sqrt(5) = sqrt(20)
hence B
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Intern  B
Joined: 24 Oct 2014
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Location: Spain
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Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)  [#permalink]

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I have a question:

It is a PS question, so the stem has to be enough to answer the question.
If ABC wasn't a right triangle, it wouldn't be possible.

Therefore, I believe checking whether there is or not a right angle is not necessary and we should jump straight to calculating the distance between (1,2) and (5,4) because it has to be the hypotenuse (ergo the diameter).

Do you agree?

Thank you.
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GMAT 1: 700 Q50 V33 GPA: 3.95
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If a circle passes through points (1, 2), (2, 5), and (5, 4)  [#permalink]

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bz9 wrote:
If a circle passes through points $$(1, 2)$$, $$(2, 5)$$, and $$(5, 4)$$, what is the diameter of the circle?

A. $$\sqrt{18}$$
B. $$\sqrt{20}$$
C. $$\sqrt{22}$$
D. $$\sqrt{26}$$
E. $$\sqrt{30}$$

$$1st Method:$$

The equation of circle with $$(h,k)$$ as centre is given by, $$(x-h)^2 + (y-k)^2 = r^2$$
As per the given equation,
$$(1-h)^2+(2-k)^2 = r^2 ---------1$$
$$(2-h)^2 + (5-k)^2 = r^2---------2$$
$$(5-h)^3 + (4-k)^2 = r^2---------3$$

Solving above 3 equations we get, $$(h,k) = (3,3)$$. Therefore, $$D = \sqrt{20}.$$

$$2nd Method:$$

The Points A, B & C form a right angle traingle. Hnece, the diameter will be equal to the distance between the Points forming the Hypotenous.
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Intern  B
Joined: 01 Feb 2018
Posts: 18
Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)  [#permalink]

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How do we exactly know that the line combining these two points go through the center? What if (in another question) it does not go through the center?
Intern  B
Joined: 02 Feb 2018
Posts: 36
Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)  [#permalink]

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Fernandocma wrote:
I have a question:

It is a PS question, so the stem has to be enough to answer the question.
If ABC wasn't a right triangle, it wouldn't be possible.

Therefore, I believe checking whether there is or not a right angle is not necessary and we should jump straight to calculating the distance between (1,2) and (5,4) because it has to be the hypotenuse (ergo the diameter).

Do you agree?

Thank you.

Agree, that's also what I thought Re: If a circle passes through points (1, 2), (2, 5), and (5, 4)   [#permalink] 11 Nov 2018, 09:35

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