Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$  represent the sides of a triangle whose centroid is G, then the equation of the circle  which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is A:  ${x^2} + {y^2} - x - y = 0$
B:  ${x^2} + {y^2} - 9x - y + 8 = 0$
C:  ${x^2} + {y^2} - 4x + 2y = 0$
D:  ${x^2} + {y^2} + 9x + y + 8 = 0$
I got the answer but had the choice not been provided how would I find the solution.
My steps are elaborated below
$3{x^2}y + 2x{y^2} - 18xy = 0 \Rightarrow xy\left( {3x + 2y - 18} \right) = 0$
The triangle is formed by line x=0, y=0 and 3x+2y-18=0
Centroid  $G = \left( {\frac{{6 + 0 + 0}}{3},\frac{{9 + 0 + 0}}{3}} \right) = \left( {2,3} \right)$
Let the centre of the circle is (h,k)
Circle passes through (2,3) and (1,1), hence the radius is
${r^2} = {\left( {h - 1} \right)^2} + {\left( {k - 1} \right)^2} = {\left( {h - 2} \right)^2} + {\left( {k - 3} \right)^2} \Rightarrow 2h + 4k = 11$
${x^2} + {y^2} + 2gx + 2fy + c + \lambda \left( {x - y + 1} \right) = 0$, represent the equation of the circle touching the line x-y+1=0 and passing through (2,3) and (1,1)
${x^2} + {y^2} + 2x\left( {g + \frac{\lambda }{2}} \right) + 2y\left( {f - \frac{\lambda }{2}} \right) + c + \lambda  = 0$
$Centre:\left( { - \left( {g - \frac{\lambda }{2}} \right), - \left( {f - \frac{\lambda }{2}} \right)} \right)$
Putting it in 2h+4k=1
$ - 2\left( {g - \frac{\lambda }{2}} \right) - 4\left( {f - \frac{\lambda }{2}} \right) = 11 \Rightarrow  - 2g - 4f + \lambda  + 2\lambda  = 11 \Rightarrow 2g + 4f = 3\lambda  - 11$
By Hit and Trial I am getting choice (B) as it satisfies $2h+4k=11$ where $h=\frac{9}{2}$ and $k=\frac{1}{2}$.
Had the option not been given how do I proceed
A: 
The centroid $G$ belongs to the line
$\mathcal{L}:$ $y=x+1$.
Since the circle $\mathcal{C}$ must pass through
$G(2,3)$ and $D(1,1)$, it must also pass through
the point $K(4,4)$,
which is a reflection of $D$ wrt the perpendicular through $G$.
Hence, the sought circle is the circumcircle of $\triangle GDK$
with the side lengths
\begin{align} 
|GD|=|GK|&=\sqrt5
,\\
|DK|&=3\sqrt2
,
\end{align}
the area
\begin{align} 
S_{GDK}&=
\tfrac12\,|DK|\,\sqrt{|GD|^2-\tfrac14\,|DK|^2}
=\tfrac32
\end{align}
and the circumradius
\begin{align} 
R&=\frac{|GD|^2|DK|}{4S_{GDK}}
=\tfrac52\,\sqrt2
.
\end{align}
Since in this special simple case
$GO$ is the diagonal
of the $\tfrac52\times\tfrac52$ square,
the coordinates of the center of the circle $\mathcal C$
can be found as
\begin{align} 
O&=G+(\tfrac52,\, -\tfrac52)
=(\tfrac92,\, \tfrac12)
.
\end{align}
In case of general $\triangle ABC$
with the side lengths $a,b,c$,
we could find
the center $O$
using a known expression
\begin{align}
O&=
\frac{a^2\,(b^2+c^2-a^2)\,A+b^2\,(a^2+c^2-b^2)\,B+c^2\,(b^2+a^2-c^2)\,C}
{a^2\,(b^2+c^2-a^2)+b^2\,(a^2+c^2-b^2)+c^2\,(b^2+a^2-c^2)}
\\
&=
\frac{a^2\,(b^2+c^2-a^2)\,A+b^2\,(a^2+c^2-b^2)\,B+c^2\,(b^2+a^2-c^2)\,C}
{16S_{ABC}^2}
,
\end{align}
which, of course, works in this special case as well
and provides the same result.
A: Hint:
$k=\dfrac{11-2h}4$
For any circle, the distance of a tangent from the center $(h,\dfrac{11-2h}4)=$radius
$$r^2=\dfrac{(h-k+1)^2}{1^2+1^2}=\dfrac{\left(h+1-\dfrac{11-2h}4\right)^2}2$$
Again $r^2=(h-1)^2+(k-1)^2=(h-1)^2+\left(\dfrac{11-2h}4-1\right)^2$
Equate the two values of $r^2$ to find $h$
A: Start with the general form of a circle:
$$x^2+y^2 + ax+by+c = 0$$
It passes through $(1, 1)$ and $(2,3)$, so $a+b+c = -2$ and $2a+3b+c = -13$. Therefore $(2a + 3b + c) - 2(a+b+c) = b-c = -9 \Rightarrow b = c - 9$.
Now, since the line $x-y+1 =0 \Rightarrow y = x+1$ is tangent to the circle, substitute in $y = x+1$.
This equation gives us the possible values of $x$, so by setting $\Delta = 0$, we ensure there is only one (real) value of $x$ common to both the circle and line, which is what a tangent is:
$$x^2 + (x+1)^2 + ax + b(x+1) + c = 0$$
$$\Rightarrow 2x^2 + (a+b+2)x + (b+c+1) = 0$$
$$\Rightarrow B^2-4AC = 0 : (a+b+2)^2 - 8(b+c+1) = 0$$
$$\Rightarrow (-c)^2 - 8(b+c+1) = 0 \tag{$a+b+c = -2$}$$
$$\Rightarrow c^2 -8(c - 9 + c + 1) = 0 \tag{$b=c-9$}$$
which gives $(a,b,c) = (-9, 1, 8)$ as expected.
A: Let $(h,k)$ be the centre then

*

*$(h,k)$ must lie on the perpendicular bisector of line joining $(2,3),(1,1)$

*$(h,k)$ must satisfy the locus given in this result
You have the ingredients to finsih..
Method -2
let circle be $S=x^2+y^2+2gx+2fy+c=0$

*

*distance of point from$(-g,-f)$ to line $x-y+1=0$ must be $\sqrt{g^2+f^2-c}$

*$(1,1),(2,3)$ must satisfy $S=0$
You have 3 equations 3 variables ...
