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Q.

The circumcentre of the triangle with vertices at A(5,12),B(12,5),C(213,313) is

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a

(1,1)

b

(0,1)

c

(0,0)

d

(1,0)

answer is C.

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Detailed Solution

Given vertices A(5,12),B(12,5),C(213,313)  let S=(x1)

SA=(x2x1)2+(y2y1)2

=(x5)2+(y12)2(1)

SB=(x2x1)2+(y2y1)2

=(x12)2+(y5)2(2)

Now equate (1)=(2)

(x5)2+(y12)2=(x12)2+(y5)2

Now squaring on both sides

(x5)2+(y12)2=(x12)2+(y5)2

x2+2510x+y2+14424y=x2+14424x+y2+2510y

14x14y=0

14x=14y

x=y

SA=(x5)2+(y12)2(3)

We know that,

Now SC=(x2x1)2+(y2y1)2

=(x213)2+(y313)2(4)

Now equate (3)=(4)

(x5)2+(y12)2=(x213)2+(y313)2

Now squaring on both sides

(x5)2+(y12)2=(x213)2+(y313)

x2+2510x+y2+14424y=x2+52413x+y2+117613y

16910x24y=169413x613y

413x10x=24y613y

Now put (1) and (2) solved equation

x=4

413y10y=24y613y

y(41310)=y(24613y)

y(4131024+613)=0

y=0(x,y)=(0,0)

Now y=0  means x=0

Then the circumcentre S(x,y)=(0,0)

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