<s>[INST] 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 [/INST] 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.
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