The Coordinates of the Center of a Circle

Oct 13, 2025 by recoveryshake.com 22 views

Introduction

In geometry, a circle is a set of points that are all equidistant from a central point known as the center. The distance between the center and any point on the circle is called the radius. The diameter of a circle is twice the length of the radius. In this problem, we are given the coordinates of the center of a circle and its diameter. We need to find the value(s) of the variable 'a' that satisfies the given conditions.

The Coordinates of the Center of a Circle

The coordinates of the center of a circle are given as ${$ (2a, a-7)$}$. This means that the x-coordinate of the center is ${ $2a\$}$ and the y-coordinate is ${$ a-7$}$.

The Diameter of the Circle

The diameter of the circle is given as ${ $10\sqrt{2}\$}$ units. Since the diameter is twice the length of the radius, we can find the radius by dividing the diameter by 2.

The Equation of a Circle

The equation of a circle with center ${$ (h, k)$}$ and radius ${$ r$}$ is given by ${(x-h)^2 + (y-k)^2 = r^2}$ . We can use this equation to find the value(s) of 'a' that satisfy the given conditions.

Substituting the Values

We know that the center of the circle is ${$ (2a, a-7)$}$ and the radius is ${ $5\sqrt{2}\$}$ . We can substitute these values into the equation of a circle:

${(x-2a)^2 + (y-(a-7))^2 = (5\sqrt{2})^2}$

Simplifying the Equation

We can simplify the equation by expanding the squares:

${(x-2a)^2 = x^2 - 4ax + 4a^2}$

${(y-(a-7))^2 = y^2 - 2ay + 49}$

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

${x^2 - 4ax + 4a^2 + y^2 - 2ay + 49 = (5\sqrt{2})^2}$

Expanding the Right-Hand Side

We can expand the right-hand side of the equation:

${(5\sqrt{2})^2 = 25 \times 2 = 50}$

So, the equation becomes:

${x^2 - 4ax + 4a^2 + y^2 - 2ay + 49 = 50}$

Simplifying the Equation

We can simplify the equation by combining like terms:

${x^2 - 4ax + 4a^2 + y^2 - 2ay = 1}$

Substituting the Point

We know that the circle passes through the point ${$ (11, -9)$}$. We can substitute these values into the equation:

${11^2 - 4a(11) + 4a^2 + (-9)^2 - 2a(-9) = 1}$

Simplifying the Equation

We can simplify the equation by expanding the squares and combining like terms:

${121 - 44a + 4a^2 + 81 + 18a = 1}$

Combining Like Terms

We can combine like terms:

${4a^2 - 26a + 200 = 1}$

Simplifying the Equation

We can simplify the equation by subtracting 1 from both sides:

${4a^2 - 26a + 199 = 0}$

Solving the Quadratic Equation

We can solve the quadratic equation using the quadratic formula:

${a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$

In this case, ${$ a = 4$, [ $b = -26\$, and \[$ c = 199$. Substituting these values into the quadratic formula, we get:

[a = \frac{-(-26) \pm \sqrt{(-26)^2 - 4(4)(199)}}{2(4)}}$

Simplifying the Expression

We can simplify the expression by evaluating the square root:

${a = \frac{26 \pm \sqrt{676 - 3168}}{8}}$

${a = \frac{26 \pm \sqrt{-2492}}{8}}$

The Solution is Not Real

Since the square root of a negative number is not real, the solution is not real.

Conclusion

In this problem, we were given the coordinates of the center of a circle and its diameter. We needed to find the value(s) of the variable 'a' that satisfies the given conditions. We used the equation of a circle to find the value(s) of 'a'. However, the solution is not real, which means that there is no value of 'a' that satisfies the given conditions.

Final Answer

The final answer is $\boxed{No real solution}$ .

Introduction

In the previous article, we discussed how to find the value(s) of the variable 'a' that satisfies the given conditions. However, the solution was not real, which means that there is no value of 'a' that satisfies the given conditions. In this article, we will answer some frequently asked questions related to the problem.

Q: What is the equation of a circle?

A: The equation of a circle with center ${$ (h, k)$}$ and radius ${$ r$}$ is given by ${(x-h)^2 + (y-k)^2 = r^2}$ .

Q: How do I find the value(s) of 'a' that satisfy the given conditions?

A: To find the value(s) of 'a', you need to substitute the values of the center and the radius into the equation of a circle. Then, you can simplify the equation and solve for 'a'.

Q: What if the solution is not real?

A: If the solution is not real, it means that there is no value of 'a' that satisfies the given conditions. In this case, you need to re-evaluate the problem and check for any errors.

Q: Can I use the quadratic formula to solve the equation?

A: Yes, you can use the quadratic formula to solve the equation. The quadratic formula is given by ${a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}$ .

Q: What is the significance of the diameter of a circle?

A: The diameter of a circle is twice the length of the radius. It is an important concept in geometry and is used to calculate the area and circumference of a circle.

Q: Can I use the equation of a circle to find the value(s) of 'a' if the circle passes through a point?

A: Yes, you can use the equation of a circle to find the value(s) of 'a' if the circle passes through a point. You need to substitute the values of the center and the point into the equation of a circle and solve for 'a'.

Q: What if the equation of a circle has no solution?

A: If the equation of a circle has no solution, it means that there is no value of 'a' that satisfies the given conditions. In this case, you need to re-evaluate the problem and check for any errors.

Q: Can I use the equation of a circle to find the value(s) of 'a' if the circle has a diameter?

A: Yes, you can use the equation of a circle to find the value(s) of 'a' if the circle has a diameter. You need to substitute the values of the center and the diameter into the equation of a circle and solve for 'a'.

Q: What is the relationship between the center and the radius of a circle?

A: The center of a circle is the point that is equidistant from all points on the circle. The radius of a circle is the distance from the center to any point on the circle.

Q: Can I use the equation of a circle to find the value(s) of 'a' if the circle has a center and a radius?

A: Yes, you can use the equation of a circle to find the value(s) of 'a' if the circle has a center and a radius. You need to substitute the values of the center and the radius into the equation of a circle and solve for 'a'.

Conclusion

In this article, we answered some frequently asked questions related to the problem of finding the value(s) of the variable 'a' that satisfies the given conditions. We discussed the equation of a circle, how to find the value(s) of 'a', and what to do if the solution is not real. We also discussed the significance of the diameter of a circle and how to use the equation of a circle to find the value(s) of 'a' if the circle passes through a point or has a diameter.

Final Answer

The final answer is $\boxed{No real solution}$ .