Find the equation of the circle which circumscribes the triangle formed by the line: `\ \ y=x+2,3y=4x\ a n d\ 2y=3x`

Explanation

Given equations of line are
y=x+2…(i)
3y=4x….(ii)
2y=3x ….(iii)
From Eqs. (i) and (ii)
`(4x)/3=x+2`
`rArr4x=3x+6`
`rArr x=6`
On putting x=6 in Eq.(i) we get
y=8
`therefore` Point,A=(6,8)
From Eqs. (i) and (iii),
`(3x)/2=x+2`
`rArr3x=2x+4rArrx=4`
When x=4 then y=6
`therefore` Point, B=(4,6)
From Eqs. (ii) and (iii) `x_(1)=0_(1),y=0`
Now, C=(0,0)
Let the equation of circle is
`x^(2)+y^(2)+2gxn+2fy+c=0`
since,the points A(6,8),B(4,6) and C(0,0) lie on this circle
36+64+12g+16f+c=0
`rArr 12g+16f+c=-100....(iv)`
and 16+36+8g+12f+c=0...(v)
`rArr 8g+12f+c=-52`....(vi)
`rArr c=0`
From Eqs. (iv) , (v) and (vi),
12g+16f=-100
`rArr3g+4f+25=0`
`rArr2g+3f+13=0`
`rArrg/(+52-75)=f/(50-39)=1/(9-8)`
`rArrg/(-23)=f/11=1/1`
`rArrg=-23,f=11`
So, the equation of the circle is
`x^(2)+y^(2)-46x+22y+0=0`
`rArrx^(2)+y^(2)-46x+22y=0`