If I have the equation $ax^2+ay^2+bx+cy+d=0$ how do I get the equation where the circumference goes through the points P = (1,1), Q = (−1,−1) and R = (−1,1)
I have it in mind to solve it with a matrix, but the instructions seem confusing to me, can I get some help?
A clever strategy (which I got from this generalization) is to simplify the following determinant equation: $$ \det\begin{bmatrix} x^2+y^2 & x & y & 1 \\ x_0^2+y_0^2 & x_0 & y_0 & 1 \\ x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ \end{bmatrix} = 0 $$ where $(x_0,y_0) = (1,1)$, $(x_1,y_1) = (-1,-1)$, and $(x_2,y_2) = (-1,1)$ are the three points.
This is the equation of a circle because the first row of the matrix is the only one that depends on $x$ and $y$; the determinant is linear in each row, so the equation will have the form $a(x^2+y^2) + bx + cy + d = 0$.
(It's possible that $a = 0$, in which case we'll get a line: a degenerate circle. But this only happens when we are given three collinear points.)
This circle passes through all three points, because when we set $x = x_i$ and $y = y_i$, the matrix has two identical rows, so it's singular and the determinant is $0$.
In this problem, by plotting the three points, it might become easy to find the circle by hand, but this method works well in general.
The given points are vertices of an isosceles right triangle,
$O(0,0)$ is the center of its hypotenuse $PQ$, and $|PQ|=2\sqrt2$.
Thus is the equation of the circumference $$x^2+y^2=2.$$
Dividing by $a\neq 0$ we obtain
$$ax^2+ay^2+bx+cy+d=0\iff x^2+y^2+\frac b a x+\frac c a y+ \frac d a=0$$
then use the three given condition to find $b/a$, $c/a$ and $d/a$.
If you need to use matrices/linear algebra, consider this approach: If you move $d$ to the other side of the equation $ax^2 + ay^2 + bx + cy + d = 0$, you get $ax^2 + ay^2 + bx + cy = -d$. We can view the left hand side of this equation as the product of two matrices: \begin{eqnarray} \left[ax^2 + ay^2 + bx + cy\right] \;=\begin{bmatrix} x^2 & y^2 & x & y \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix} \end{eqnarray}
Since $ax^2 + ay^2 + bx + cy = -d$, we get the matrix equation \begin{eqnarray} \begin{bmatrix} x^2 & y^2 & x & y \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix} = \left[-d\right]. \end{eqnarray}
Now we have 3 points that satisfy this matrix equation. Taking the first point, $P=(1,1)$, and plugging it into this equation, we obtain \begin{eqnarray} \begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix} = \left[-d\right]. \end{eqnarray} Similary, we plug in $Q=(-1, -1)$ to get \begin{eqnarray} \begin{bmatrix} 1 & 1 & -1 & -1 \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix} = \left[-d\right] \end{eqnarray} and $R=(-1,1)$ to get \begin{eqnarray} \begin{bmatrix} 1 & 1 & -1 & 1 \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix} = \left[-d\right]. \end{eqnarray} Now we have 3 $simultaneous$ $linear$ $equations$, which can be combined into the matrix equation \begin{eqnarray} \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & 1 & -1 & 1 \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix} = \begin{bmatrix}-d \\ -d \\ -d \end{bmatrix}. \end{eqnarray} From here, one can use standard Gaussian Elimination to obtain \begin{eqnarray} \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\cdot \begin{bmatrix}a \\a \\ b \\ c\end{bmatrix} = \begin{bmatrix}-d \\ 0 \\ 0 \end{bmatrix}. \end{eqnarray} This tells us that $b=c=0$, and $2a = -d$. Coming back to our original equation we insert these values to obtain \begin{equation} ax^2 + ay^2 + 0 x + 0 y = 2a \end{equation} We can divide by $a$ since $a$ must be nonzero (otherwise we don't have a circle!) to obtain the final equation \begin{equation} x^2 + y^2 = 2.\end{equation}
$ax^2+a y^2+bx+cy+d=0$ is the equation of an ellipse, not a line You can note that it seems all P,Q and R are at a distance of $\sqrt{2}$ of the center $(0,0)$. Therefore, let's imagine an ellipse centered around $(0,0)$, it has the following equation: $(X-0)^2 + a(Y-0)^2 = b$. Therefore, $b=c=0$.The coefficient in front of $X^2$ and $Y^2$ is the same, you can simplify (it's non-zero). $X^2 + Y^2 = -d/a$ Plug it in $P = (1,1)$: $1 + 1 = 2 = -d/a$ so $a=-d/2$ The final equation becomes: $X^2 + Y^2 = -d/a = 2$
