Question

Which of the following is a possible equation for a circle that is tangent to both the \(x\)-axis and the line \(x=4\) ? A) \((x+2)^2+(y+2)^2=4\) B) \((x+2)^2+(y-2)^2=4\) C) \((x-2)^2+(y+4)^2=4\) D) \((x-6)^2+(y-2)^2=4\)

Short answer

The center of the circle lies between the x-axis and the line x=4 and has the coordinates \((4 - r, r)\), where r is the radius. The general equation for a circle is \((x-a)^2 + (y-b)^2 = r^2\). Our derived equation, using the center coordinates \((4-r, r)\), is \((x-(4-r))^2 + (y-r)^2 = r^2\). Upon comparing this equation with the given answer choices, none of them represent a circle that is tangent to both the x-axis and the line x=4.

Step-by-step solution

Determine the center and radius of the circle.

Since the circle is tangent to both the x-axis and the line x=4, we can assume that the center of the circle lies between these two lines. Using this assumption, we can deduce the coordinates of the center of the circle (a, b) to be \((4 - r, r)\), where r denotes the radius of the circle. This is because the distance from the center to x=4 equals the radius r and the distance from the center to the x-axis is also equal to the radius r.

Recall the standard equation for a circle.

The equation of a circle with center \((a, b)\) and radius r is given by the formula: \((x-a)^2 + (y-b)^2 = r^2\)

Apply the center and radius to the equation.

Using the center coordinates \((4-r, r)\), replace a and b in the standard equation of a circle and obtain the following equation: \((x-(4-r))^2 + (y-r)^2 = r^2\)

Compare the circle equations with answer choices.

Now, let's look at the answer choices and compare each equation to our derived equation: A) \((x+2)^2+(y+2)^2=4\): Center is at \((-2, -2)\) with radius 2, so it's incorrect. B) \((x+2)^2+(y-2)^2=4\): Center is at \((-2, 2)\) with radius 2, so it's also incorrect. C) \((x-2)^2+(y+4)^2=4\): Center is at (2, -4) with radius 2, so it's incorrect. D) \((x-6)^2+(y-2)^2=4\): Center is at \((6, 2)\) with radius 2. Since the center is at (6, 2), we have \(4 - r = 6\) (by comparing with center \((4 - r, r)\)), which implies that the radius r = -2. But the radius cannot be negative, so we know that none of the answer choices provided represent a circle that is tangent to both the x-axis and the line x=4.

Key concepts

Tangent to x-axis

Understanding the concept of a circle being tangent to the x-axis is crucial for solving geometry problems proficiently. Imagine a flat surface where the x-axis runs horizontally. A circle tangent to the x-axis lightly touches it at exactly one point. This makes the distance from the circle's center to the x-axis equal to the radius of the circle. In terms of coordinates, if a circle is tangent to the x-axis at the point \( (a,0) \), then the center of the circle would be at \( (a, r) \) where \( r \) is the radius of the circle. This concept helps in visualizing the position of the circle in a two-dimensional plane and is pivotal in identifying the correct equation for the circle based on its position relative to the axis.

For instance, in the problem provided, knowing that the circle is tangent to the x-axis allows us to conclude that the y-coordinate of the circle’s center must be the same as the radius. This knowledge is instrumental in narrowing down the possible equations for the circle.

Circle Standard Equation

A circle's standard equation plays a definitive role in solving circle-related geometry problems. It is given by the formula \[ (x-a)^2 + (y-b)^2 = r^2 \] where \( a \) and \( b \) are the x and y coordinates of the circle's center, and \( r \) is the radius. This equation is derived from the Pythagorean theorem and represents every point \( (x, y) \) lying on the circumference of the circle. In summary, the standard equation forms the backbone for addressing any circle's geometry and helps in deducing properties such as the center, radius, and the relationship of the circle to other geometric figures like lines and axes.

Geometry SAT Problems

Geometry SAT problems often challenge students to apply their knowledge of shapes, sizes, and the properties of space. When faced with such problems, it’s essential to approach them step by step, breaking down the question into known principles and formulas. Circles, being one of the fundamental geometric shapes, usually require students to be familiar with concepts like tangents, chords, secants, and their standard equations. In addition, spatial reasoning is required to visualize and solve these problems efficiently. By consistently practicing problems that involve the properties of circles, such as the one presented in the text, students can hone their skills and become adept at handling SAT geometry questions.

Circle Center and Radius

Determining the center and radius of a circle is essential in geometry. The center-point \( (a, b) \) gives us the fixed point from which every point on the circumference is equidistant, and this equidistant measure is the circle's radius \( r \). Together, these two parameters not only define a circle uniquely but also impact its position in relation to other geometric figures. For example, when a circle is tangent to the x-axis, the y-coordinate of its center equals its radius, as is the case in the given exercise. It’s important to grasp this relationship since misidentifying the circle’s center or radius can lead to incorrect solutions or misconceptions about the circle’s spatial orientation within a coordinate system.