Question

Find the circumcenter whose sides are x+y+2=0, 5x-y-2=0, x-2y+5

Answer 1

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Let

$AB→x+y+2=0 \: \: \: \: \: \: \: (1)$

$BC→5x−y−2=0 \: \: \: \: \: \: \: \: \: (2)$

$CA→x−2y+5=0 \: \: \: \: \: \: \: (3)$

My plan is to find a find a pair of perpendiculars at each end of a side and take the average which will be perpendicular bisector of the side and proceed with the intersection of two such bisectors.

(A) Perpendicular bisector of AB.

Perpendicular to AB will be of the form

$x−y+…=0 \: \: \: \: \: \: \: \: \: (4)$

Equation of any line passing through B will be linear combination of (1) and (2)

$λ(x+y+2)+μ(5x−y−2)=0$

$⟹(λ+5μ)x+(λ−μ)y+2λ−2μ=0 \: \: \: \: \: \: \: \: \: \: \: \: \: (5)$

This should be compared with (4)

$\frac{λ+5μ}{1} = \frac{λ−μ}{ - 1} ⟹$

$λ=−2μChoose \: \: \: \: \: \: μ=1, λ=−2λ$

Back substitution in (5) gives a line perpendicular to AB at B.

$x−y−2=0 \: \: \: \: \: \: \: \: \: \: \: (6)$

Equation of any line passing through AA will be linear combination of (1)(1) and (3)(3)

$λ(x+y+2)+μ(x−2y+5)=0$

$⟹(λ+μ)x+(λ−2μ)y+2λ+5μ=0 \: \: \: \: \: \: \: \: \: \: \: \: \: (7)$

This should be compared with (4)

$\frac{λ+μ}{1}= \frac{λ−2μ}{ - 1}⟹$

$2λ=μChoose \: \: \: \: \: μ=2, λ=12λ$

Back substitution in (7) gives a line perpendicular to AB at A.

$x−y+4=0(8)$

The perpendicular bisector of ABAB is the average of (6) and (8)

$x−y+1=0 \: \: \: \: \: \: \: \: \: \: \: (9)$

(B) Perpendicular bisector of BC.

Following the same procedure as in (A) we can obtain this as

$x+5y−3=0 \: \: \: \: \: \: \: \: \: \: (10)$

Simultaneous solution of (9) and (10) gives

$Circumcenter \: = - \frac {1}{3} , \frac{2}{3}$

However the equation of circum circle can not be easily determined by this method.