Problem 39 - Algebra 2

Question

Write the equation of the circle that passes through the given point and has a center at the origin. (Hint: You can use the distance formula to find the radius.) $$ (4,-3) $$

Step-by-Step Solution

Answer

The equation of the circle is \(x^2 + y^2 = 25\)

Step 1: Identifying the known variables

We have a circle with center at the origin (0,0). The point (4,-3) lies on the circle. Therefore, we know the coordinates for h and k to be 0, and have a point on the circle by which to determine the radius.

Step 2: Calculating the radius

Since the radius is the distance from the center of the circle to any point on the circle, we will apply the distance formula (or Pythagorean theorem in this case) to find the radius (r). The distance between points (x1, y1) and (x2, y2) is given by \( \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \). Let's consider (x1, y1) as the center (0,0) and (x2, y2) as the given point (4, -3). So, the radius will be \( \sqrt{(4 - 0)^2 + ((-3) - 0)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \). Therefore, the radius of the circle is 5.

Step 3: Formulating the equation

Now that we have the radius and the coordinates of the center, we'll substitute them into the standard equation of a circle, \((x - h)^2 + (y - k)^2 = r^2\). Thus, the equation of the circle is \((x - 0)^2 + (y - 0)^2 = 5^2\), which simplifies to \(x^2 + y^2 = 25\).

Key Concepts

Distance Formula

The Distance Formula is a crucial concept in geometry used to calculate the distance between two points in a two-dimensional space. This formula is derived from the Pythagorean Theorem. It can be expressed as:

• Given two points ewline ewline li>\((x_1, y_1)\) and \((x_2, y_2)\), the distance \(d\) between them is given by:

\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]This formula can be easily remembered as an extension of the Pythagorean Theorem, where the differences \((x_2 - x_1)\) and \((y_2 - y_1)\) represent the lengths of the legs of a right-angled triangle, with \(d\) being the hypotenuse.
By applying this to find the radius of a circle whose center is at the origin and passes through a given point, we establish the radius as the distance from the origin \((0,0)\) to that point.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry, especially pertinent in scenarios involving right triangles. The theorem states that for any right triangle, the square of the length of the hypotenuse \(c\) (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides, \(a\) and \(b\):

• \(a^2 + b^2 = c^2\).

This relationship helps in deducing various geometric measurements, such as the radius of a circle when you know the coordinates of a point on the circle.
_Subsection

• Using it to find the radius: When dealing with a circle centered at the origin \((0,0)\), the radius \(r\) can be perceived as the hypotenuse of the right triangle formed by the x and y axes.
• In our exercise, point \((4, -3)\) forms this triangle, calculating the radius as \(r = \sqrt{4^2 + (-3)^2} = 5\).

This helps to bridge understanding between basic algebraic formulas and geometric visualization, facilitating more robust comprehension of geometrical circles.

Standard Form of a Circle Equation

The standard form of a circle equation is a key tool that allows us to precisely define the location and size of a circle in the Cartesian coordinate system. The standard equation is given by:

• \((x - h)^2 + (y - k)^2 = r^2\)

here \((h, k)\) denotes the center of the circle and \(r\) represents the radius. This equation succinctly encodes the geometry of a circle.
_{Adapting to the Origin}

• When the circle's center is at the origin \((0,0)\), the equation simplifies to \(x^2 + y^2 = r^2\).
• This form is intuitive because it shows that any point \((x, y)\) on the circle satisfies a constant distance \(r\) from the origin.
• In our exercise, knowing the radius is 5, the equation simplifies further to \(x^2 + y^2 = 25\), encapsulating all points that lie on this circle.

Understanding this form reinforces the interconnectedness of algebra and geometry and exemplifies the elegance of mathematical expressions.

Origin as Center

Assigning the origin as the center in geometrical problems simplifies many calculations and represents one of the most basic components of analytic geometry. When the center of a circle is at the origin, it acts as the most central point in the Cartesian plane

• The coordinates \((0,0)\) remove potential translation terms from equations, yielding a cleaner, simpler form.
• This central point means any calculated radius is merely the distance from the origin to any point on the circle.

_{Practical Implication}

• For our equation, \((x - 0)^2 + (y - 0)^2 = r^2\) simplifies to \(x^2 + y^2 = r^2\).
• This form is direct and easy to apply for quick calculations.
• The clean nature of having the origin as the center makes it ideal for introductory problems where mastering the basic principles of circles and distances is paramount.

Through this setup, students gain clarity in visualization and computation, providing a foundation for solving more complex geometric tasks.