If a circle passes through points (1, 2), (2, 5), and (5, 4)

Question

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}

Bunuel · Accepted Answer

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. 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.

Bunuel · Answer

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} Look at the diagram below: https://gmatclub.com/forum/download/file.php?id=20109 Calculate the lengths of the sides of triangle ABC: AB=\sqrt{10} ; BC=\sqrt{10} ; AC=\sqrt{20}=\sqrt{2}*\sqrt{10} ; As we see the ratio of the sides of triangle ABC is 1:1:\sqrt{2} , so ABC is 45°-45°-90° right triangle (in 45°-45°-90° right triangle the sides are always in the ratio 1:1:\sqrt{2} ). So, we have right triangle ABC inscribed in the circle. Now, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle , so AC=diameter=\sqrt{20} . Answer: B. Inscribed triangle.png

Fig · Answer

(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)

lonewolf · Answer

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?

gottabwise · Answer

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.

pramoddikshith · Answer

I dont know how you could assuse AC is the diameter. Because there is nothing to indicate that AC goes through the center right? I am a little confused about the solution.

Bunuel · Answer

AC is the diameter because angle at B is 90 degrees. Check this post: circle-passes-through-points-1-2-2-5-and-42105.html#p672194

daboo343 · Answer

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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If a circle passes through points (1, 2), (2, 5), and (5, 4)

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