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

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

Explanation for the correct options:

Step 1: Finding the point of intersection

The point of intersection of $AB$ and $AC$ is point $A$.

Solving $x+2y=5$ and $x+y=6$

$$\begin{gathered} \Rightarrow y=-1;x=7 \\ \therefore A\equiv \left(7,-1\right) \end{gathered}$$

Similarly, the point of intersection of $AB$ and $BC$ is point $B$.

Solving equations $x+2y=5$ and $2x+y=4$ we get

$$\begin{gathered} \Rightarrow y=2;x=1 \\ \therefore B\equiv \left(1,2\right) \end{gathered}$$

The point of intersection of $AC$ and $BC$ is point $C$.

Solving equations $x+y=6$ and $2x+y=4$ we get

$$\begin{gathered} \Rightarrow y=8;x=-2 \\ \therefore C\equiv \left(-2,8\right) \end{gathered}$$

Step 2: Finding the center and radius

Now, center of the circle is $O\left(a,b\right)$

We know that, $A{O}^2=B{O}^2=C{O}^2$$\Rightarrow {\left(7-a\right)}^2+{\left(-1-b\right)}^2={\left(1-a\right)}^2+{\left(2-b\right)}^2={\left(-2-a\right)}^2+{\left(8-b\right)}^2$

Evaluating these three equations, we get

$a=\frac{17}{2},b=\frac{19}{2}$.

Hence, center of the circle is $\left(a,b\right)=\left(\frac{17}{2},\frac{19}{2}\right)$.

$$\begin{gathered} \text{Radius}=l\left(OA\right) \\ =\sqrt{{\left(a-1\right)}^2+{\left(b-2\right)}^2} \\ =\sqrt{{\left(\frac{17}{2}-1\right)}^2+{\left(\frac{19}{2}-2\right)}^2} \\ =\sqrt{{\left(\frac{15}{2}\right)}^2+{\left(\frac{15}{2}\right)}^2} \\ =\sqrt{2}\left(\frac{15}{2}\right) \\ =\frac{15}{\sqrt{2}} \end{gathered}$$

Step 3: Finding the equation of circle

Now the equation of circle is

$$\begin{gathered} \Rightarrow {\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2 \\ \Rightarrow {\left(x-\frac{17}{2}\right)}^2+{\left(y-\frac{19}{2}\right)}^2={\left(\frac{15}{\sqrt{2}}\right)}^2 \\ \Rightarrow x^2-17x+\frac{289}{4}+y^2-19y+\frac{361}{4}=\frac{450}{4} \\ \Rightarrow x^2+y^2-17x-19y+\frac{289+361-450}{4}=0 \\ \Rightarrow x^2+y^2-17x-19y+50=0 \end{gathered}$$

Therefore, option (D) is the correct answer.