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Chapter 10: Problem 72

What is the radius of the circle with equation \((x-2)^{2}+3+(y+1)^{2}=7 ?\)

Short Answer

Expert verified
The radius of the circle with the corrected equation \((x-2)^{2}+(y+1)^{2}=7\) is \(\sqrt{7}\).

Step by step solution

01

Recognize the Circle Equation

Recognize that the given equation, \((x-2)^{2}+3+(y+1)^{2}=7\), is a circle equation on the form \((x-h)^{2}+(y-k)^{2}=r^{2}\). But there is a mistake in given equation it should be \((x-2)^{2}+(y+1)^{2}=7\) as the constant term (3 in this case) shouldn't be in equation of a circle.
02

Identify \(r^{2}\)

On the right side of the equation, there is number 7 which represent \(r^{2}\).
03

Compute for the Radius

To find the circle's radius, determine the square root of the \(r^{2}\) value (7 in this case). Calculate the square root of 7.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Radius Calculation
Calculating the radius of a circle is a straightforward process once you've identified the value of \( r^2 \) in the circle equation. The radius is a measure of the distance from the center of the circle to any point on its circumference. To find the radius from an equation of the form \((x-h)^2 + (y-k)^2 = r^2\), follow these steps:
  • First, ensure that your equation matches the standard circle equation form and isolate \( r^2 \) on one side of the equation.
  • For instance, in the equation \((x-2)^2 + (y+1)^2 = 7\), \( r^2 \) is clearly 7.
  • Calculate the square root of \( r^2 \) to find \( r \), the radius. Thus, the radius \( r \) is \( \sqrt{7} \).
Radius calculation boils down to finding the square root of the number equated to \( r^2 \). Always remember this key point: the square root of \( r^2 \) gives you the radius.
Equation of a Circle
Understanding the equation of a circle is fundamental when you're dealing with circle-related math problems. A circle's equation in the standard format looks like \((x-h)^2 + (y-k)^2 = r^2\). Here are the key components:
  • \( (h, k) \) represents the center of the circle. The values of \( h \) and \( k \) shift the circle around on the coordinate plane.
  • \( r^2 \) is the radius squared. It determines the size of the circle.
In our problem, the given equation is \((x-2)^2 + (y+1)^2 = 7\). By comparing this with the standard form, it's clear:
  • The center of the circle is \( (2, -1) \).
  • \( r^2 = 7 \), which means the radius \( r \) is \( \sqrt{7} \).
Note how the equation reveals both location and size of the circle. Mastering this form allows you to write and interpret any circle's equation effortlessly.
Algebraic Manipulation
Often, algebraic manipulation is necessary to transform equations into a familiar form, like the standard circle equation. Here鈥檚 a simple approach to ensure your circle's equation is set correctly:
  • Start by identifying discrepancies or extraneous terms that should not be present in the circle equation. In the example, the initial equation had an extra constant term (3).
  • Remove or adjust these terms to match the standard form \((x-h)^2 + (y-k)^2 = r^2\).
  • Rearrange the terms if necessary to isolate the complete squares for both \( x \) and \( y \).
These steps often involve:
  • Expanding and combining like terms.
  • Completing the square, if the equation isn't neatly presented.
Effective algebraic manipulation is crucial for simplifying equations and solving problems accurately by reformatting them for easy computation.

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Most popular questions from this chapter

The foci of a hyperbola are \((0,-4)\) and \((0,8) .\) Which additional information would allow you to write an equation for the hyperbola? F. location of the center G. location of one vertex H. midpoint of transverse axis J. distance from the center to a focus

Simplify each expression. \(\ln e\)

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Identify the conic section represented by each equation by writing the equation in standard form. For a parabola, give the vertex. For a circle, give the center and the radius. For an ellipse or a hyperbola, give the center and the foci. Sketch the graph. \(x^{2}+4 y^{2}-2 x-15=0\)
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