1. Interpret the given geometry

The triangle has two sides lying on the lines $y=0$ and $x+y+1=0.$

So these two sides intersect at one vertex of the triangle.

Let

• $A$ be the intersection of $y=0$ and $x+y+1=0$.

From $y=0$, the second line gives $x+0+1=0 \Rightarrow x=-1.$ So $A=(-1,0).$

Let the other two vertices be:

• $B=(b,0)$ on the $x$-axis,
• $C=(c,-c-1)$ on the line $x+y+1=0$.

The orthocenter is given as $H=(1,1).$

2. Use the fact that altitudes pass through the orthocenter

Altitude from $B$

Side $AC$ lies on the line $x+y+1=0,$ which has slope $-1$. So the altitude from $B$ must have slope $1$.

Equation of altitude through $B=(b,0)$: $y=x-b.$ Since $H=(1,1)$ lies on it, $1=1-b \Rightarrow b=0.$ Thus $B=(0,0).$

Altitude from $C$

Side $AB$ lies on the $x$-axis, i.e. $y=0$. An altitude perpendicular to the $x$-axis is vertical. So altitude from $C$ is $x=c.$ Since $H=(1,1)$ lies on it, $c=1.$ Thus $C=(1,-2).$

So the triangle vertices are $A=(-1,0),\quad B=(0,0),\quad C=(1,-2).$

3. Find the circle through these three points

Let the circle be $x^2+y^2+Dx+Ey+F=0.$

Substitute the three points.

Using $B=(0,0)$

$F=0.$

Using $A=(-1,0)$

$(-1)^2+0+D(-1)+0+0=0$ $1-D=0 \Rightarrow D=1.$

Using $C=(1,-2)$

$1+4+D(1)+E(-2)+0=0$ $5+1-2E=0$ $6-2E=0 \Rightarrow E=3.$

Therefore the circle is $x^2+y^2+x+3y=0.$

4. Match with the options

This corresponds to:

Option B: $x^2+y^2+x+3y=0.$

5. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So they agree.