Which Of The Following Circles Lie Completely Within The Fourth Quadrant? Check All That Apply.A. \[$(x-5)^2 + (y+5)^2 = 9\$\]B. \[$(x-9)^2 + (y+9)^2 = 16\$\]C. \[$(x-2)^2 + (y+7)^2 = 64\$\]D. \[$(x-12)^2 + (y+0)^2 = 72\$\]

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Which of the Following Circles Lie Completely Within the Fourth Quadrant? Check All That Apply

The fourth quadrant is one of the four quadrants in the Cartesian coordinate system. It is defined by the x-axis and the y-axis, where the x-axis is the horizontal axis and the y-axis is the vertical axis. The fourth quadrant is located in the bottom-right section of the coordinate plane, where both the x and y coordinates are negative.

Understanding the Fourth Quadrant

To determine which circles lie completely within the fourth quadrant, we need to analyze each equation and find the x and y intercepts. The x-intercept is the point where the circle intersects the x-axis, and the y-intercept is the point where the circle intersects the y-axis.

Analyzing Equation A

The first equation is {(x-5)^2 + (y+5)^2 = 9$}$. To find the x and y intercepts, we can set y = 0 and x = 0, respectively.

  • Setting y = 0: {(x-5)^2 + (0+5)^2 = 9$ [(x−5)2+25=9$\[(x-5)^2 + 25 = 9\$ \[(x-5)^2 = -16$ Since the square of any real number cannot be negative, this equation has no real solutions. Therefore, the circle does not intersect the x-axis.

  • Setting x = 0: [$(0-5)^2 + (y+5)^2 = 9$ [25+(y+5)2=9$\[25 + (y+5)^2 = 9\$ \[(y+5)^2 = -16$ Since the square of any real number cannot be negative, this equation has no real solutions. Therefore, the circle does not intersect the y-axis.

Analyzing Equation B

The second equation is [(x-9)^2 + (y+9)^2 = 16\$}. To find the x and y intercepts, we can set y = 0 and x = 0, respectively.

  • Setting y = 0: {(x-9)^2 + (0+9)^2 = 16$ [(x−9)2+81=16$\[(x-9)^2 + 81 = 16\$ \[(x-9)^2 = -65$ Since the square of any real number cannot be negative, this equation has no real solutions. Therefore, the circle does not intersect the x-axis.

  • Setting x = 0: [$(0-9)^2 + (y+9)^2 = 16$ [81+(y+9)2=16$\[81 + (y+9)^2 = 16\$ \[(y+9)^2 = -65$ Since the square of any real number cannot be negative, this equation has no real solutions. Therefore, the circle does not intersect the y-axis.

Analyzing Equation C

The third equation is [(x-2)^2 + (y+7)^2 = 64\$}. To find the x and y intercepts, we can set y = 0 and x = 0, respectively.

  • Setting y = 0: {(x-2)^2 + (0+7)^2 = 64$ [(x−2)2+49=64$\[(x-2)^2 + 49 = 64\$ \[(x-2)^2 = 15$ [x−2=±15$\[x-2 = \pm\sqrt{15}\$ \[x = 2 \pm \sqrt{15}$ Since the x-coordinate is between -2 and 2, the circle does not intersect the x-axis.

  • Setting x = 0: [$(0-2)^2 + (y+7)^2 = 64$ [4+(y+7)2=64$\[4 + (y+7)^2 = 64\$ \[(y+7)^2 = 60$ [y+7=±60$\[y+7 = \pm\sqrt{60}\$ \[y = -7 \pm \sqrt{60}$ Since the y-coordinate is between -7 and 0, the circle intersects the y-axis.

Analyzing Equation D

The fourth equation is [(x-12)^2 + (y+0)^2 = 72\$}. To find the x and y intercepts, we can set y = 0 and x = 0, respectively.

  • Setting y = 0: [(x−12)2+(0+0)2=72$\[(x-12)^2 + (0+0)^2 = 72\$ \[(x-12)^2 = 72$ [x−12=±72$\[x-12 = \pm\sqrt{72}\$ \[x = 12 \pm \sqrt{72}$ Since the x-coordinate is between 12 and 0, the circle intersects the x-axis.

  • Setting x = 0: [$(0-12)^2 + (y+0)^2 = 72$ [144+(y+0)2=72$\[144 + (y+0)^2 = 72\$ \[(y+0)^2 = -72$ Since the square of any real number cannot be negative, this equation has no real solutions. Therefore, the circle does not intersect the y-axis.

Conclusion

Based on the analysis of each equation, we can conclude that:

  • Equation A does not intersect the x or y axis, so it does not lie completely within the fourth quadrant.
  • Equation B does not intersect the x or y axis, so it does not lie completely within the fourth quadrant.
  • Equation C intersects the y-axis, but not the x-axis, so it partially lies within the fourth quadrant.
  • Equation D intersects the x-axis, but not the y-axis, so it partially lies within the fourth quadrant.

Therefore, the only circle that lies completely within the fourth quadrant is None of the above.
Q&A: Understanding Circles in the Fourth Quadrant

In our previous article, we analyzed four different equations of circles and determined which ones lie completely within the fourth quadrant. However, we received many questions from readers who wanted a more in-depth explanation of the concepts involved. In this article, we will answer some of the most frequently asked questions about circles in the fourth quadrant.

Q: What is the fourth quadrant?

A: The fourth quadrant is one of the four quadrants in the Cartesian coordinate system. It is defined by the x-axis and the y-axis, where the x-axis is the horizontal axis and the y-axis is the vertical axis. The fourth quadrant is located in the bottom-right section of the coordinate plane, where both the x and y coordinates are negative.

Q: How do I determine if a circle lies within the fourth quadrant?

A: To determine if a circle lies within the fourth quadrant, you need to analyze the equation of the circle and find the x and y intercepts. The x-intercept is the point where the circle intersects the x-axis, and the y-intercept is the point where the circle intersects the y-axis. If both intercepts are negative, then the circle lies within the fourth quadrant.

Q: What if the circle intersects the x-axis or y-axis, but not both?

A: If the circle intersects the x-axis or y-axis, but not both, then it partially lies within the fourth quadrant. This means that the circle has a portion of its circumference within the fourth quadrant, but not the entire circle.

Q: Can a circle have a negative radius?

A: No, a circle cannot have a negative radius. The radius of a circle is a measure of its distance from the center to the edge, and it is always a positive value.

Q: How do I find the x and y intercepts of a circle?

A: To find the x and y intercepts of a circle, you can set y = 0 and x = 0, respectively, in the equation of the circle. This will give you the x and y intercepts, which you can then use to determine if the circle lies within the fourth quadrant.

Q: What if the equation of the circle is not in the standard form (x-h)^2 + (y-k)^2 = r^2?

A: If the equation of the circle is not in the standard form, you can still find the x and y intercepts by completing the square or using other algebraic techniques. However, be careful to ensure that you are working with the correct equation and that you are not introducing any errors.

Q: Can a circle have a negative center?

A: No, a circle cannot have a negative center. The center of a circle is a point in the coordinate plane that is equidistant from all points on the circle, and it is always a positive value.

Q: How do I determine if a circle is tangent to the x-axis or y-axis?

A: To determine if a circle is tangent to the x-axis or y-axis, you need to find the x and y intercepts of the circle and check if they are equal to zero. If the x-intercept is zero, then the circle is tangent to the x-axis. If the y-intercept is zero, then the circle is tangent to the y-axis.

Q: Can a circle have a negative radius and a negative center?

A: No, a circle cannot have a negative radius and a negative center. The radius and center of a circle are always positive values.

Conclusion

In this article, we have answered some of the most frequently asked questions about circles in the fourth quadrant. We hope that this information has been helpful in clarifying the concepts involved and has provided a better understanding of the subject. If you have any further questions or need additional clarification, please don't hesitate to ask.