Question

Find the equation of the circle circumscribing the triangle formed by the lines $x+y=0,2x+y=4$ and $x+2y=5$

• A. $x^2+y^2+17x+19y+50=0$
• B. $x^2+y^2+17x-11y+30=0$
• C. $x^2+y^2-17x-11y+30=0$
• D. $x^2+y^2-17x-19y+50=0$

Answer 1

The correct option is D

$x^2+y^2-17x-19y+50=0$

Family of circles circumscribing a triangle whose sides are given by $L_1=0,L_2=0$and $L_3=0$ is

$L_1{L}_2+\lambda L_2{L}_3+\mu L_3{L}_1=0$

$(x+y-6)(2x+y-4)+\lambda (2x+y-4)(x+2y-5)+\mu (x+2y-5)(x+y-6)$

$(2x^2+xy-4x+2xy+y^2-4y-12x-6y+24)$

$+\lambda (2x^2+4xy-10x+xy+2y^2-5y-4x-8y+20)$

$+\mu (x^2+xy-6x+2y^2-12y-5x-5y+30)$

$=(2+2\lambda +\mu )x^2+(1+2\lambda +2\mu )y^2+(3+5\lambda +3\mu )xy+(-16-14\lambda -11\mu )x+(-10-13\lambda -17\mu )y+(24+20\lambda +30\mu )=0$- - - - - - (1)

Since, it is an equation of circle

So coefficient of xy should be zero and coefficient of $x^2=$coefficient of $y^2$

Coefficient of xy = zero

So, $3+5\lambda +3\mu =0$

$5\lambda +3\mu =-3$- - - - - - (2)

Coefficient of $x^2$ = coefficient of $y^2$

$2+2\lambda +\mu =1+2\lambda +2\mu$

$\mu =1$

Substituting $\mu$ in equation 2

$5\lambda +3=-3$

$\lambda =\frac{-6}{5}$

Now, substituting the values of $\lambda \&\mu$ in equation (1)

$(2+2\left(\frac{-6}{5}\right)+1)x^2+(1+2\left(\frac{-6}{5}\right)+2\times 1)y^2+0xy$

$-16+14\left(\frac{-6}{5}\right)+11\times 1)x-(10+13\left(\frac{-6}{5}\right)+17\times 1)y+(24+20\left(\frac{-6}{5}\right)+30\times 1)$

$\frac{3}{5}{x}^2+\frac{3}{5}{y}^2-\frac{51}{5}\times \frac{57}{5}y+30=0$

Multiplying both sides by $\frac{5}{3}$

$x^2+y^2-17x-19y+50=0$

so, equation of circle circumscribing triangle is

$x^2+y^2-17x-19y+50=0$