IEB 2009 Matric Paper: Geometry of Touching Circles and Triangle Area Maximization

Problem Overview and Identification of Circle A

This material originates from the 2009 Independent Examinations Board (IEB) matric paper, focusing on analytical geometry and trigonometry within the context of intersecting or touching circles. The problem defines three circles with centers $A$, $B$, and $C$ that touch each other. Central to the geometric setup is Circle $A$, which is defined by the standard form equation $(x-2)^2 + (y+1)^2 = 4$. From this equation, we can determine the coordinates of the center and the radius of the circle. The center of Circle $A$ is located at point $(2, -1)$, and the radius $r_A$ is the square root of the constant term on the right side of the equation, yielding $r_A = \text{2 units}$.

Determination of Center and Equation of Circle C

The coordinates and properties of Circle $C$ are derived using the geometric relationship between centers $A$ and $C$. It is explicitly stated that the line segment $AC$ is parallel to the $y$-axis and has a length of $3$ units. Because $AC$ is vertical, the $x$-coordinate of $C$ must match the $x$-coordinate of $A$, which is $2$. Depending on the orientation (upward from $A$), the $y$-coordinate of $C$ is found by adding the length of $AC$ to the $y$-coordinate of $A$, resulting in $-1 + 3 = 2$. Thus, the center of Circle $C$ is $(2, 2)$.

To find the radius of Circle $C$, we use the fact that Circle $A$ and Circle $C$ touch each other. The distance between the centers is given as $AC = 3$. Since the circles touch externally, this distance is the sum of their radii: $AC = r_A + r_C$. Substituting the known values, we have $3 = 2 + r_C$, which simple subtraction shows $r_C = \text{1 unit}$. The standard equation for Circle $Center\text{ }C$ is therefore $(x-2)^2 + (y-2)^2 = 1^2$, which simplifies to $(x-2)^2 + (y-2)^2 = 1$.

Area of Triangle ABC in Terms of the Variable e

The problem introduces a triangle formed by the centers of the three circles, denoted as $\Delta ABC$. The area of any triangle can be calculated using the trigonometric area formula: $\text{Area} = \frac{1}{2} \times \text{side}_1 \times \text{side}_2 \times \text{sin}(\text{included angle})$. In this scenario, the sides adjacent to the angle at center $A$ are given as $AC = 3$ and $AB = 6$. The transcript specifies the angle $C\hat{A}B$ using the variables "e" and "€". Using the variable "e" as defined in part (b), the expression for the area of $\Delta ABC$ becomes:

$\text{Area} = \frac{1}{2} \times 3 \times 6 \times \sin(e)$

$\text{Area} = 9\sin(e)$

This expression represents the area of the triangle as a function of the angle located at vertex $A$.

Maximization of Triangle Area within the Specified Range

Part (c) of the problem asks for the specific value of the angle variable, denoted here as "€", in the interval "€ ∈ [0^{\circ}; 180^{\circ}]" that will maximize the area of $\Delta ABC$. Based on the area function derived previously, $\text{Area} = 9\sin(€)$, the area is maximized when the sine of the angle is at its peak value. In the given range of $0^{\circ}$ to $180^{\circ}$, the sine function reaches its maximum value of $1$ when the angle is exactly $90^{\circ}$. Therefore, the area is maximized when the triangle is a right-angled triangle with the right angle at vertex $A$.

$\text{Maximum Area} = 9\sin(90^{\circ}) = 9 \times 1 = \text{9 square units}$

Equation of Circle Center B at Maximum Area

To find the equation of Circle $B$ when the area is maximized, we first determine its radius and then its center coordinates. Since Circle $A$ and Circle $B$ touch each other and the distance $AB$ is given as $6$, the radius of Circle $B$ is calculated as $r_B = AB - r_A$. This results in $r_B = 6 - 2 = \text{4 units}$.

When the area is at its maximum, the angle at $A$ is $90^{\circ}$. Since $AC$ is parallel to the $y$-axis (vertical), the segment $AB$ must be parallel to the $x$-axis (horizontal) to maintain the perpendicular relationship. Starting from center $A(2, -1)$ and moving horizontally by the distance $AB = 6$, the location of point $B$ could be $(2+6, -1)$ or $(2-6, -1)$. This gives two possible centers: $(8, -1)$ or $(-4, -1)$.

For center $B(8, -1)$, the equation of the circle is $(x-8)^2 + (y+1)^2 = 16$. For center $B(-4, -1)$, the equation is $(x+4)^2 + (y+1)^2 = 16$. These equations satisfy the condition that Circle $B$ has a radius of $4$ and is positioned such that $AB$ is perpendicular to $AC$.