Recall The Equation For A Circle With Center { (h, K)$}$ And Radius { R$}$. At What Point In The First Quadrant Does The Line With Equation { Y = 2x + 3$}$ Intersect The Circle With Radius 6 And Center [$(0,

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Introduction

In mathematics, the intersection of a circle and a line is a fundamental concept that has numerous applications in various fields, including geometry, trigonometry, and calculus. In this article, we will explore the equation of a circle with a given center and radius, and then find the point of intersection with a line in the first quadrant.

Recall 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$}$

This equation represents a circle centered at {(h, k)$}$ with a radius of {r$}$. The equation is derived from the distance formula, which states that the distance between two points {(x_1, y_1)$}$ and {(x_2, y_2)$}$ is given by:

{d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$}$

In the case of a circle, the distance between the center and any point on the circle is equal to the radius.

Given Circle and Line

We are given a circle with a radius of 6 and center {(0, 0)$}$. The equation of the circle is:

{x^2 + y^2 = 36$}$

We are also given a line with the equation {y = 2x + 3$}$. To find the point of intersection between the circle and the line, we need to substitute the equation of the line into the equation of the circle.

Substituting the Equation of the Line

Substituting {y = 2x + 3$}$ into the equation of the circle, we get:

{x^2 + (2x + 3)^2 = 36$}$

Expanding the equation, we get:

{x^2 + 4x^2 + 12x + 9 = 36$}$

Combine like terms:

${5x^2 + 12x - 27 = 0\$}

This is a quadratic equation in the form {ax^2 + bx + c = 0$}$. We can solve this equation using the quadratic formula:

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

In this case, {a = 5$}$, {b = 12$}$, and {c = -27$}$. Plugging these values into the quadratic formula, we get:

{x = \frac{-12 \pm \sqrt{12^2 - 4(5)(-27)}}{2(5)}$}$

Simplifying the equation, we get:

{x = \frac{-12 \pm \sqrt{144 + 540}}{10}$}$

{x = \frac{-12 \pm \sqrt{684}}{10}$}$

{x = \frac{-12 \pm 26.16}{10}$}$

We have two possible values for {x$}$:

{x_1 = \frac{-12 + 26.16}{10} = 1.416$}$

{x_2 = \frac{-12 - 26.16}{10} = -3.816$}$

Since we are looking for the point of intersection in the first quadrant, we discard the negative value of {x$}$ and use the positive value:

{x = 1.416$}$

Now that we have the value of {x$}$, we can substitute it into the equation of the line to find the value of {y$}$:

{y = 2x + 3$}$

{y = 2(1.416) + 3$}$

{y = 2.832 + 3$}$

{y = 5.832$}$

Therefore, the point of intersection between the circle and the line in the first quadrant is {(1.416, 5.832)$}$.

Conclusion

In this article, we have shown how to find the point of intersection between a circle and a line in the first quadrant. We started by recalling the equation of a circle with a given center and radius, and then substituted the equation of the line into the equation of the circle. We solved the resulting quadratic equation using the quadratic formula and found the values of {x$}$ and {y$}$ that correspond to the point of intersection. The point of intersection is {(1.416, 5.832)$}$.

References

  • [1] "Circle" by Math Open Reference. Retrieved 2023-02-20.
  • [2] "Line" by Math Open Reference. Retrieved 2023-02-20.
  • [3] "Quadratic Formula" by Math Is Fun. Retrieved 2023-02-20.

Further Reading

  • [1] "Geometry" by Khan Academy. Retrieved 2023-02-20.
  • [2] "Trigonometry" by Khan Academy. Retrieved 2023-02-20.
  • [3] "Calculus" by Khan Academy. Retrieved 2023-02-20.
    Q&A: Intersection of a Circle and a Line =============================================

Introduction

In our previous article, we explored the equation of a circle with a given center and radius, and then found the point of intersection with a line in the first quadrant. In this article, we will answer some frequently asked questions (FAQs) related to the intersection of a circle and a line.

Q: What is the equation of a circle with center (h, k) and radius r?

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 point of intersection between a circle and a line?

A: To find the point of intersection between a circle and a line, you need to substitute the equation of the line into the equation of the circle. This will result in a quadratic equation, which you can solve using the quadratic formula.

Q: What is the quadratic formula?

A: The quadratic formula is a mathematical formula that is used to solve quadratic equations. It is given by:

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

Q: How do I use the quadratic formula to solve a quadratic equation?

A: To use the quadratic formula, you need to identify the values of {a$}$, {b$}$, and {c$}$ in the quadratic equation. Then, you can plug these values into the quadratic formula and simplify the expression to find the value of {x$}$.

Q: What if the quadratic equation has no real solutions?

A: If the quadratic equation has no real solutions, it means that the line does not intersect the circle. In this case, the quadratic equation will have complex solutions, which are not real numbers.

Q: Can I use the quadratic formula to solve a quadratic equation with complex solutions?

A: Yes, you can use the quadratic formula to solve a quadratic equation with complex solutions. However, the solutions will be complex numbers, which are not real numbers.

Q: How do I find the point of intersection between a circle and a line in the second quadrant?

A: To find the point of intersection between a circle and a line in the second quadrant, you need to follow the same steps as before. However, you will need to use the negative value of {x$}$ and the positive value of {y$}$.

Q: Can I use the quadratic formula to solve a quadratic equation with a negative value of x?

A: Yes, you can use the quadratic formula to solve a quadratic equation with a negative value of {x$}$. However, you will need to use the negative value of {x$}$ and the positive value of {y$}$.

Q: What if the line is tangent to the circle?

A: If the line is tangent to the circle, it means that the line touches the circle at a single point. In this case, the quadratic equation will have a single solution, which is the point of tangency.

Q: Can I use the quadratic formula to solve a quadratic equation with a single solution?

A: Yes, you can use the quadratic formula to solve a quadratic equation with a single solution. However, the solution will be a single value of {x$}$, which is the point of tangency.

Conclusion

In this article, we have answered some frequently asked questions (FAQs) related to the intersection of a circle and a line. We have covered topics such as the equation of a circle, the quadratic formula, and the point of intersection between a circle and a line. We hope that this article has been helpful in answering your questions and providing you with a better understanding of the intersection of a circle and a line.

References

  • [1] "Circle" by Math Open Reference. Retrieved 2023-02-20.
  • [2] "Line" by Math Open Reference. Retrieved 2023-02-20.
  • [3] "Quadratic Formula" by Math Is Fun. Retrieved 2023-02-20.

Further Reading

  • [1] "Geometry" by Khan Academy. Retrieved 2023-02-20.
  • [2] "Trigonometry" by Khan Academy. Retrieved 2023-02-20.
  • [3] "Calculus" by Khan Academy. Retrieved 2023-02-20.