Problem 71
Question
Find the equation of the circle. Center (3,3)\(;\) passes through the origin.
Step-by-Step Solution
Verified Answer
Answer: The equation of the circle is (x - 3)^2 + (y - 3)^2 = 18.
1Step 1: Determine the radius using the distance formula
We need to find the distance between the center of the circle (3, 3) and the origin (0, 0). This distance is equal to the radius of the circle.
The distance formula is given by:
$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
Plugging in the coordinates:
$$r = \sqrt{(0 - 3)^2 + (0 - 3)^2}$$
2Step 2: Calculate the radius
Now, we must simplify the expression to find the length of the radius:
$$r = \sqrt{(-3)^2 + (-3)^2}$$
$$r = \sqrt{9 + 9}$$
$$r = \sqrt{18}$$
3Step 3: Write the equation of the circle
The standard equation of a circle is given by:
$$(x - a)^2 + (y - b)^2 = r^2$$
Where (a, b) is the center of the circle and r is the radius. Since we already have the center (3, 3) and the radius \(\sqrt{18}\), we can plug these values into the equation:
$$(x - 3)^2 + (y - 3)^2 = (\sqrt{18})^2$$
4Step 4: Simplify the equation
Finally, let's simplify the expression by squaring the radius:
$$(x - 3)^2 + (y - 3)^2 = 18$$
So, the equation of the circle is \((x - 3)^2 + (y - 3)^2 = 18\).
Key Concepts
Distance FormulaRadius CalculationStandard Equation of a Circle
Distance Formula
When we're tasked with finding the equation of a circle, one of the most critical components we need is the circle's radius. The distance formula comes to the rescue as it allows us to calculate the straight-line distance between two points in a coordinate plane, which, in the context of a circle, translates into figuring out the radius.
The formula is rooted in the Pythagorean theorem and is represented as: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points. For a circle, one of the points will be the center, and the other will usually be a point on the circumference. This is the starting point for determining every circle’s size and position within the coordinate plane.
The formula is rooted in the Pythagorean theorem and is represented as: \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] where \((x_1, y_1)\) and \((x_2, y_2)\) are the coordinates of the two points. For a circle, one of the points will be the center, and the other will usually be a point on the circumference. This is the starting point for determining every circle’s size and position within the coordinate plane.
Radius Calculation
The radius of a circle is essentially half its diameter, and it's the distance from the center of the circle to any point on its edge. In practical terms, if you know two points on the circumference, or one point on it and the center, you can calculate the radius with the distance formula mentioned earlier.
Applying the Distance Formula to Radius Calculation
In the given exercise, we have the center of the circle at \((3, 3)\) and a point on the circumference, the origin \((0, 0)\). Plugging these into the distance formula yields \[r = \sqrt{(0 - 3)^2 + (0 - 3)^2}\] After simplifying, we find that \[r = \sqrt{18}\], which we can express as \(3\sqrt{2}\) for a more precise radius calculation.Standard Equation of a Circle
Armed with the knowledge of the radius and the coordinates of the center, we can now write down the circle's equation. The standard format for the equation of a circle centered at a point \((a, b)\) with radius \(r\) is represented as: \[ (x - a)^2 + (y - b)^2 = r^2 \]
For our example, the center is \((3, 3)\) and the radius squared is \(18\), leading us to the equation: \[ (x - 3)^2 + (y - 3)^2 = 18 \] This equation encompasses every point \((x, y)\) that lies on the circle and is essential for understanding a circle's properties in algebraic geometry. Through practice, one can easily learn to visualize and plot circles by applying this elegant mathematical concept.
For our example, the center is \((3, 3)\) and the radius squared is \(18\), leading us to the equation: \[ (x - 3)^2 + (y - 3)^2 = 18 \] This equation encompasses every point \((x, y)\) that lies on the circle and is essential for understanding a circle's properties in algebraic geometry. Through practice, one can easily learn to visualize and plot circles by applying this elegant mathematical concept.
Other exercises in this chapter
Problem 70
Simplify, and write the given number without using absolute values. $$|(-2) 3|$$
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Solve the equation and check your answers. $$1-\frac{3}{x}=\frac{40}{x^{2}}$$
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Simplify, and write the given number without using absolute values. $$3-|2-5|$$
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Solve the equation and check your answers. $$\frac{2}{x^{2}}-\frac{5}{x}=4$$
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