The equation of the circumcircle of the triangle formed by the lines whose combined equation is given by (x+y-4) (xy-2x-y+2)=0, is

Options

A. `x^(2)+y^(2)-5x-3y+8=0`
B. `x^(2)+y^(2)-3x-5y+8=0`
C. `x^(2)+y^(2)-3x-5y-8=0`
D. none of these

Answer

B

Explanation

The combined equation of the sides of the triangle is
`(x+y-4) (x-1)(y-2)=0 rArr x+y-4=0, x=1, y=2`
The equation of a second degree curve passing through the vertices of the triangle is
`lambda(x+y-4) (x-1)+mu (x-1)(y-2)+v(y-2)(x+y-4)=0 " " ...(i)`
The equation will represent a circle, if
Coeff. of `x^(2)` = Coeff. of `y^(2)` and Coeff. of xy=0
`rArr lambda = v and lambda + mu +v=0`
`rArr lambda = v and mu = -2v`
Substituting these values in (i), we obtain
`(x+y-4)(x-1)-2(x-1)(y-2)(x+y-4)=0`
`rArr x^(2)+y^(2)-3x-5y+8=0`
This is the equation of the required circumcircle .
`ul("ALITER")` The equations of the sides of the triangle are x+y-4=0, x=1 and y=2. Clearly, they form a right angled triangle. Therefore, circumcentre is the mid-point of hypotenuse and circumradius is half of the length of the hypotenuse. The coordinates of the circumcentre is (3/2, 5/2) and circumradius `=(sqrt(2))/(2)=(1)/(sqrt(2))`

So, the equation of the circumcircle is
`(x-(3)/(2))^(2)+(y-(5)/(2))^(2)=((1)/(sqrt(2)))^(2) rArr x^(2)+y^(2)-3x-5y+8=0`.