IMO B
let the centre be x,y
eqn 1(dist b/n center and point 1,2) is x
x^2+1-2x+y^2+4-4y
similarly eqn2 x^2+y^2+25+4-10y-4x
eqn3 x^2+y^2+25+16-10x-8y

by equating eqn 1 and 2 we get x+3y=12
and by equating eqn2 and 3 we get 3x-y=6solving x=y=3
radius=sqrt5 and dia=sqrt20

getso 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)

couldn't find a shorter way, but to solve the equations.

IMO B.

Anyone with a short-cut ?

the three points (1, 2)(2,5),and (5,4) form a right angle triangle

distance between (1, 2)(2,5)= V10
distance between (2,5)(5,4)= V10
distance between (1, 2)(5,4)= V20
hence diameter = length of hypotenuse =sqrt(20)

option B

If you plot the point on a graph
A (1,2) , B ( 2,5) and C ( 5,4) it becomes clear AB is perpendicular to BC.
Not convinced, slope of AB*slope of BC = -1
(5-2)/(2-1) x(5-4)/(2-5)=-1
Since the circle touch all the points of the triangle ABC and <ABC is 90°
AC has to be the diameter.
AC² = (5-1)²+(4-2)² = 20
IMO B.

Frankly if ABC is not a rt. angle triangle finding the diameter will test concepts beyond the scope of GMAT.

liferocks wrote:the three points (1, 2)(2,5),and (5,4) form a right angle triangle

distance between (1, 2)(2,5)= V10
distance between (2,5)(5,4)= V10
distance between (1, 2)(5,4)= V20
hence diameter = length of hypotenuse =sqrt(20)

option B

naaaaaaice! thanks buddy!

liferocks wrote:the three points (1, 2)(2,5),and (5,4) form a right angle triangle

distance between (1, 2)(2,5)= V10
distance between (2,5)(5,4)= V10
distance between (1, 2)(5,4)= V20
hence diameter = length of hypotenuse =sqrt(20)

option B

https://www.codecogs.com/reference/maths ... circle.php


Guys, i think are we writing olympiads....?? hehehe!

kstv wrote:If you plot the point on a graph
A (1,2) , B ( 2,5) and C ( 5,4) it becomes clear AB is perpendicular to BC.
Not convinced, slope of AB*slope of BC = -1
(5-2)/(2-1) x(5-4)/(2-5)=-1
Since the circle touch all the points of the triangle ABC and <ABC is 90°
AC has to be the diameter.
AC² = (5-1)²+(4-2)² = 20
IMO B.

Frankly if ABC is not a rt. angle triangle finding the diameter will test concepts beyond the scope of GMAT.

Thanks for the wonderful explaination..

kstv wrote:If you plot the point on a graph
A (1,2) , B ( 2,5) and C ( 5,4) it becomes clear AB is perpendicular to BC.
Not convinced, slope of AB*slope of BC = -1
(5-2)/(2-1) x(5-4)/(2-5)=-1
Since the circle touch all the points of the triangle ABC and <ABC is 90°
AC has to be the diameter.
AC² = (5-1)²+(4-2)² = 20
IMO B.

Frankly if ABC is not a rt. angle triangle finding the diameter will test concepts beyond the scope of GMAT.

nice one..