Question

The equation of the circle circumscribing the triangle formed by the lines $x+y=6$, $2x+y=4$ and $x+2y=5$ is:

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

Answer 1

The correct option is D $x^2+y^2-17x-19y+50=0$

Solving given three line the vertices of the triangle are $A(1,2),B(-2,8)$ and $C(7,-1)$

And let $P(a,b)$ the center of the circle.

Therefore,

$P{A}^2=P{B}^2=P{C}^2$

$\Rightarrow (a-1{)}^2+(b-2{)}^2=(a+2{)}^2+(b-8{)}^2=(a-7{)}^2+(b+1{)}^2$

Solving above equations we get $a=\frac{17}{2}$ and $b=\frac{19}{2}$

Now radius of the circle is $PA=\sqrt{(a-1{)}^2+(b-2{)}^2}=\sqrt{\frac{{15}^2}{4}+\frac{{15}^2}{4}}$

Hence equation of required circle is,

${(x-\frac{17}{2})}^2+{(y-\frac{19}{2})}^2=P{A}^2=\frac{{15}^2}{4}+\frac{{15}^2}{4}$

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