Question

The graph of $$ a\left(x^{2}+y^{2}\right)+b x+c y+d=0 $$ is a circle if \(a \neq 0\). Show that there is a circle through any three points in the plane that are not all on a line.

Short answer

A unique circle can pass through any three non-collinear points in a plane.

Step-by-step solution

Understand the Equation of a Circle

The general form of a circle in the plane \[a(x^2 + y^2) + bx + cy + d = 0\]is a quadratic equation in \(x\) and \(y\) where \(a eq 0\). This matches the equation of a circle, since rearranging \[a(x^2 + y^2) + bx + cy + d = 0\]can be restructured to the form \((x - h)^2 + (y - k)^2 = r^2\), identifying it as a circle. For this form, \(a\) typically equates to \(1\) to match standard equations.

Determine Circle Conditions

To assert that a circle can pass through any three points that are not collinear, we recognize that three non-collinear points determine exactly one unique circle. This is because such points create a triangle with a well-defined circumcircle.

Use Three Non-Collinear Points

Consider three points \((x_1, y_1), (x_2, y_2)\), and \((x_3, y_3)\). Substitute into the general form of the circle equation, leading to a system of equations:\[\begin{aligned} a(x_1^2 + y_1^2) + b x_1 + c y_1 + d &= 0, \ a(x_2^2 + y_2^2) + b x_2 + c y_2 + d &= 0, \ a(x_3^2 + y_3^2) + b x_3 + c y_3 + d &= 0.\end{aligned}\]

Solve the System of Equations

To confirm a circle through the three points, solve this system of three linear equations for the unknowns \(a, b, c, d\). The determinant of the coefficients matrix must be non-zero for a unique solution, ensuring that \(a eq 0\). Non-collinearity of points means the determinant is indeed non-zero.

Verify Existence of a Unique Solution

Check that the determinant of the matrix formed by coefficients of \(b, c, d\) is non-zero. If so, a unique relationship between \(a, b, c, d\) exists, allowing the reformation into a standard circle equation to confirm a circle through the three points.

Key concepts

Quadratic Equation

In the realm of mathematics, a quadratic equation is a type of polynomial equation of the second degree. Generally, it is expressed in the form \(ax^2 + bx + c = 0\). When it comes to circles in a coordinate plane, the quadratic equation takes a slightly different form: \(a(x^2 + y^2) + bx + cy + d = 0\). Here, the variables \(x\) and \(y\) represent the coordinates in the Cartesian plane.
If you look closely, you'll see that this is still fundamentally a quadratic equation due to the term \(x^2 + y^2\). For the equation to represent a circle, a must not be equal to zero (\(a eq 0\)).
This equation can be manipulated to fit the standard circle equation \((x - h)^2 + (y - k)^2 = r^2\), which reveals the circle's center \((h,k)\) and radius \(r\). Understanding this transformation helps us see how most circles are just cleverly disguised quadratic equations.

Circumcircle

A circumcircle is a special circle that passes through all vertices of a polygon, typically a triangle in this context. This circle is significant because it is the only circle that can encompass the triangle perfectly.
When we talk about circle conditions for any three points, we're actually considering a situation where these points form a triangle. Any set of three non-collinear points defines a unique circumcircle.
Such a circle is formed because the perpendicular bisectors of each side of the triangle meet at a single point, known as the circumcenter. This point is equidistant from all vertices of the triangle, making it the center of the circumcircle. The existence of this circle underscores how three non-collinear points are essential for determining a unique circle in geometry.

Non-collinear Points

Non-collinear points are points that do not all lie on the same straight line. In a two-dimensional plane, this means that the points can form a polygon, usually a triangle.
In the context of determining a circle, non-collinear points are crucial. If three points were collinear, they would lie on a single straight line and no unique circle could pass through all of them. This is because a straight line could technically be part of infinitely many circles.
Thus, ensuring that the points are non-collinear allows for the unique solution to the system of equations derived from substituting the points into the circle's general quadratic equation. This solution is ensured by the non-zero determinant condition, verifying that these points indeed define a specific circle.