Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point

Question

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

Answer 1

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.

Answer 2

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 ...

Answer 3

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 $h$

Answer 4

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.