Isolating Y Squared In The Equation (x+4)^2 + Y^2 = 22

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In this article, we will delve into the process of isolating y2y^2 in the given equation, (x+4)2+y2=22(x+4)^2 + y^2 = 22. This is a fundamental algebraic manipulation skill that is essential for solving various mathematical problems, especially those involving circles and other conic sections. Understanding how to isolate variables allows us to rewrite equations in different forms, making them easier to analyze and solve. We will break down each step with clear explanations and examples to ensure you grasp the concept thoroughly. Let's embark on this mathematical journey together!

Understanding the Equation

Before we begin isolating y2y^2, it's crucial to understand the structure of the equation (x+4)2+y2=22(x+4)^2 + y^2 = 22. This equation represents a circle in the Cartesian coordinate system. The general form of the equation of a circle is (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center of the circle and rr is the radius. In our case, we have (x+4)2+y2=22(x + 4)^2 + y^2 = 22, which can be rewritten as (x−(−4))2+(y−0)2=22(x - (-4))^2 + (y - 0)^2 = 22. This tells us that the center of the circle is at (−4,0)(-4, 0) and the radius squared, r2r^2, is 22.

Recognizing this structure helps us understand the context of the problem. Isolating y2y^2 might be a step towards finding the explicit form of yy as a function of xx, which can be useful for graphing the circle or solving related problems. The process involves algebraic manipulation to rearrange the equation while maintaining its balance. We will focus on applying the correct operations to both sides of the equation to achieve our goal.

Step-by-Step Solution

To isolate y2y^2 in the equation (x+4)2+y2=22(x+4)^2 + y^2 = 22, we need to eliminate the term (x+4)2(x+4)^2 from the left side of the equation. This can be achieved by subtracting (x+4)2(x+4)^2 from both sides of the equation. This maintains the equality and moves us closer to our goal.

Here are the steps:

  1. Start with the original equation: (x+4)2+y2=22(x+4)^2 + y^2 = 22
  2. Subtract (x+4)2(x+4)^2 from both sides: (x+4)2+y2−(x+4)2=22−(x+4)2(x+4)^2 + y^2 - (x+4)^2 = 22 - (x+4)^2
  3. Simplify the left side: y2=22−(x+4)2y^2 = 22 - (x+4)^2

Now, we have isolated y2y^2 on the left side. The next step is to expand the term (x+4)2(x+4)^2 on the right side to simplify the expression further. Expanding this term will involve using the binomial expansion formula or simply multiplying (x+4)(x+4) by itself.

Expanding (x+4)2(x+4)^2

Expanding (x+4)2(x+4)^2 is a crucial step in simplifying the equation. To expand this term, we can use the formula (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2. In our case, a=xa = x and b=4b = 4. Applying this formula:

(x+4)2=x2+2(x)(4)+42(x+4)^2 = x^2 + 2(x)(4) + 4^2 (x+4)2=x2+8x+16(x+4)^2 = x^2 + 8x + 16

Now that we have expanded (x+4)2(x+4)^2, we can substitute this back into our equation to further simplify the expression for y2y^2. This substitution will allow us to combine like terms and obtain a more concise form of the equation.

Substituting and Simplifying

Now that we have expanded (x+4)2(x+4)^2 as x2+8x+16x^2 + 8x + 16, we can substitute this back into the equation we derived earlier:

y2=22−(x+4)2y^2 = 22 - (x+4)^2

Substituting the expanded form:

y2=22−(x2+8x+16)y^2 = 22 - (x^2 + 8x + 16)

Next, we distribute the negative sign across the terms inside the parentheses:

y2=22−x2−8x−16y^2 = 22 - x^2 - 8x - 16

Finally, we combine the constant terms (22 and -16):

y2=−x2−8x+6y^2 = -x^2 - 8x + 6

This is the simplified form of the equation with y2y^2 isolated. It shows the relationship between y2y^2 and xx in the context of the given circle equation.

Analyzing the Result

The final result of isolating y2y^2 in the equation (x+4)2+y2=22(x+4)^2 + y^2 = 22 is:

y2=−x2−8x+6y^2 = -x^2 - 8x + 6

This equation represents y2y^2 as a quadratic function of xx. We can observe that the coefficient of x2x^2 is negative, which indicates that the parabola opens downwards if we were to graph this equation. However, it's important to remember that this equation represents the square of yy, not yy itself.

To find yy, we would need to take the square root of both sides, which would introduce both positive and negative roots, reflecting the circular nature of the original equation. The expression −x2−8x+6-x^2 - 8x + 6 provides insight into the possible values of xx for which yy is a real number. Since y2y^2 must be non-negative, we have:

−x2−8x+6≥0-x^2 - 8x + 6 ≥ 0

Solving this inequality would give us the range of xx values for which real solutions for yy exist. This kind of analysis is crucial in understanding the properties and behavior of the original circle equation.

Identifying the Correct Option

Given the options provided:

A. y2=−x2−8x+16y^2 = -x^2 - 8x + 16 B. y2=x2−8x+6y^2 = x^2 - 8x + 6 C. y2=−x2−8x+6y^2 = -x^2 - 8x + 6 D. y2=22−x2y^2 = 22 - x^2

Comparing our derived equation y2=−x2−8x+6y^2 = -x^2 - 8x + 6 with the options, we can see that option C matches our result.

Therefore, the correct answer is:

C. y2=−x2−8x+6y^2 = -x^2 - 8x + 6

This confirms that we have correctly isolated and simplified the equation to find the expression for y2y^2.

Conclusion

In conclusion, we have successfully isolated y2y^2 in the equation (x+4)2+y2=22(x+4)^2 + y^2 = 22. The step-by-step process involved subtracting (x+4)2(x+4)^2 from both sides, expanding the resulting expression, and simplifying to obtain the final equation y2=−x2−8x+6y^2 = -x^2 - 8x + 6. This process highlights the importance of algebraic manipulation skills in solving mathematical problems.

Understanding how to isolate variables is a fundamental concept that extends beyond this specific example. It is a skill that is crucial in various areas of mathematics and other scientific disciplines. By mastering these techniques, you can approach more complex problems with confidence and clarity. Remember to always focus on maintaining the balance of the equation and applying the correct operations to achieve your goal.

The correct option, as we identified, is C. y2=−x2−8x+6y^2 = -x^2 - 8x + 6. This exercise demonstrates a practical application of algebraic principles and reinforces the importance of careful and methodical problem-solving.