Which Equation Represents A Circle With A Center At { (-3, -5)$}$ And A Radius Of 6 Units?A. { (x-3)^2 + (y-5)^2 = 6$}$B. { (x-3)^2 + (y-5)^2 = 36$}$C. { (x+3)^2 + (y+5)^2 = 6$} D . \[ D. \[ D . \[ (x+3)^2 + (y+5)^2 =

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Circles are fundamental geometric shapes that play a crucial role in mathematics, particularly in geometry and trigonometry. A circle is defined as the set of all points in a plane that are equidistant from a fixed point, known as the center. In this article, we will explore the equation of a circle and how to identify the correct equation for a given circle.

The General Equation of a Circle

The general equation of a circle with center (h,k){(h, k)} and radius r{r} is given by:

(x−h)2+(y−k)2=r2{(x - h)^2 + (y - k)^2 = r^2}

This equation represents a circle with center (h,k){(h, k)} and radius r{r}. The center of the circle is the point (h,k){(h, k)}, and the radius is the distance from the center to any point on the circle.

Identifying the Center and Radius

To identify the center and radius of a circle, we need to look at the equation and extract the values of h{h}, k{k}, and r{r}. In the given equation, (x−h)2+(y−k)2=r2{(x - h)^2 + (y - k)^2 = r^2}, the values of h{h} and k{k} are the coordinates of the center, and the value of r{r} is the square of the radius.

The Given Circle

The given circle has a center at (−3,−5){(-3, -5)} and a radius of 6 units. We need to find the equation of this circle using the general equation of a circle.

Substituting the Values

To find the equation of the given circle, we need to substitute the values of h{h}, k{k}, and r{r} into the general equation of a circle. The center of the circle is (−3,−5){(-3, -5)}, so we have h=−3{h = -3} and k=−5{k = -5}. The radius of the circle is 6 units, so we have r=6{r = 6}.

Substituting these values into the general equation of a circle, we get:

(x−(−3))2+(y−(−5))2=62{(x - (-3))^2 + (y - (-5))^2 = 6^2}

Simplifying the equation, we get:

(x+3)2+(y+5)2=36{(x + 3)^2 + (y + 5)^2 = 36}

Comparing the Options

Now that we have the equation of the given circle, we can compare it with the options provided.

Option A: (x−3)2+(y−5)2=6{(x-3)^2 + (y-5)^2 = 6}

This option is incorrect because the radius is not squared.

Option B: (x−3)2+(y−5)2=36{(x-3)^2 + (y-5)^2 = 36}

This option is correct because the radius is squared.

Option C: (x+3)2+(y+5)2=6{(x+3)^2 + (y+5)^2 = 6}

This option is incorrect because the radius is not squared.

Option D: (x+3)2+(y+5)2=36{(x+3)^2 + (y+5)^2 = 36}

This option is correct because the radius is squared.

Conclusion

In conclusion, the correct equation of the circle with a center at (−3,−5){(-3, -5)} and a radius of 6 units is:

(x+3)2+(y+5)2=36{(x+3)^2 + (y+5)^2 = 36}

This equation represents a circle with center (−3,−5){(-3, -5)} and radius 6 units.

Final Answer

The final answer is:

In the previous article, we explored the equation of a circle and how to identify the correct equation for a given circle. In this article, we will answer some frequently asked questions about circle equations.

Q: What is the general equation of a circle?

A: The general equation of a circle with center (h,k){(h, k)} and radius r{r} is given by:

(x−h)2+(y−k)2=r2{(x - h)^2 + (y - k)^2 = r^2}

Q: How do I identify the center and radius of a circle from its equation?

A: To identify the center and radius of a circle from its equation, you need to look at the equation and extract the values of h{h}, k{k}, and r{r}. In the given equation, (x−h)2+(y−k)2=r2{(x - h)^2 + (y - k)^2 = r^2}, the values of h{h} and k{k} are the coordinates of the center, and the value of r{r} is the square of the radius.

Q: What is the difference between the equation of a circle and the equation of a sphere?

A: The equation of a circle is given by (x−h)2+(y−k)2=r2{(x - h)^2 + (y - k)^2 = r^2}, while the equation of a sphere is given by (x−h)2+(y−k)2+(z−l)2=r2{(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2}. The main difference is that the equation of a sphere includes an additional term for the z-coordinate.

Q: Can I have a circle with a negative radius?

A: No, a circle cannot have a negative radius. The radius of a circle is a measure of its size, and it cannot be negative.

Q: How do I find the equation of a circle with a given center and radius?

A: To find the equation of a circle with a given center and radius, you need to substitute the values of h{h}, k{k}, and r{r} into the general equation of a circle. For example, if the center of the circle is (3,4){(3, 4)} and the radius is 5 units, the equation of the circle would be:

(x−3)2+(y−4)2=52{(x - 3)^2 + (y - 4)^2 = 5^2}

Q: Can I have a circle with a center at the origin?

A: Yes, you can have a circle with a center at the origin. In this case, the equation of the circle would be:

x2+y2=r2{x^2 + y^2 = r^2}

Q: How do I find the equation of a circle with a given diameter?

A: To find the equation of a circle with a given diameter, you need to divide the diameter by 2 to get the radius, and then substitute the values of h{h}, k{k}, and r{r} into the general equation of a circle.

Q: Can I have a circle with a center at a point with coordinates (0,0){(0, 0)} and a radius of 0?

A: Yes, you can have a circle with a center at a point with coordinates (0,0){(0, 0)} and a radius of 0. In this case, the equation of the circle would be:

x2+y2=0{x^2 + y^2 = 0}

This equation represents a single point, which is the center of the circle.

Conclusion

In conclusion, the equation of a circle is a powerful tool for describing and analyzing circles. By understanding the general equation of a circle and how to identify the center and radius of a circle from its equation, you can solve a wide range of problems involving circles.

Final Answer

The final answer is:

(x−h)2+(y−k)2=r2{(x - h)^2 + (y - k)^2 = r^2}