# A-Level Maths: 10 MINUTE TOPIC REVISION: Circles (Coordinate Geometry)

TLDRThis 10-minute A-Level Maths revision video focuses on the topic of circles in coordinate geometry. It explains how to derive the equation of a circle from its standard form, including finding the center and radius, and how to complete the square if necessary. The video also covers the concept that the radius is perpendicular to the tangent of a circle, which is crucial for finding the equation of tangents and normals. Additionally, it discusses the property that the perpendicular bisector of a chord passes through the circle's center, demonstrating its application with an example involving the shortest distance from the circle's center to a chord.

### Takeaways

- π The standard form of a circle's equation allows you to quickly identify the center and radius.
- π If the circle's equation isn't in standard form, complete the square to find the center and radius.
- π’ For example, the equation x^2 + 14x + y^2 - 12y = -11 simplifies to a circle with center at (-7, 6) and radius 5.
- βοΈ Remember that the radius squared (r^2) is the value on the right side of the equation, not the radius itself.
- π The radius of a circle is perpendicular to the tangent line at the point of tangency.
- π To find the gradient of the tangent, use the negative reciprocal of the radius's gradient.
- π Use the point-slope form to write the equation of the tangent line once you have the gradient and a point of tangency.
- πΆ The perpendicular bisector of a chord passes through the center of the circle.
- π The shortest distance from the center of a circle to a chord is along the perpendicular bisector.
- π§© Use right-angle triangles within the circle to apply the Pythagorean theorem and find unknown distances.
- π Given the circle's equation, you can determine the radius and subsequently use it to find distances with Pythagoras.

### Q & A

### What is the standard form of a circle's equation?

-The standard form of a circle's equation is (x - h)Β² + (y - k)Β² = rΒ², where (h, k) is the center of the circle and r is the radius.

### How can you determine the center and radius of a circle from its equation if it's not in the standard form?

-If the equation is not in standard form, you can determine the center and radius by completing the square for both x and y terms.

### What is the process of completing the square for the x terms in the circle's equation?

-To complete the square for the x terms, take half of the coefficient of the x term, square it, and add and subtract this value to maintain the equation's balance.

### Can you provide an example of how to find the center and radius of a circle given the equation xΒ² + 14x + yΒ² - 12y = 25?

-For the equation xΒ² + 14x + yΒ² - 12y = 25, complete the square for x and y. The center is at (-7, 6) and the radius is 5, since the constant term on the right side of the equation is 25, which is 5Β².

### Why is the radius of a circle perpendicular to the tangent at any point on the circle?

-The radius is perpendicular to the tangent at any point on the circle because the tangent is a line that just touches the circle at that point, and the radius forms a right angle with it.

### How do you find the equation of a tangent to a circle given a point on the circle?

-To find the equation of a tangent, you need the gradient of the radius at the given point and then use the fact that the tangent's gradient is the negative reciprocal of the radius' gradient. Then use the point-slope form of a line with the point on the circle and the tangent's gradient.

### What is the significance of the perpendicular bisector of a chord in relation to the circle's center?

-The perpendicular bisector of a chord passes through the center of the circle. This property is useful for finding the shortest distance from the center to the chord, as it creates right angle triangles within the circle.

### How can you use the perpendicular bisector of a chord to find the shortest distance from the center to the chord?

-Since the perpendicular bisector of a chord goes through the center and bisects the chord, you can create right angle triangles within the circle. By knowing the radius and half the chord length, you can use the Pythagorean theorem to find the shortest distance.

### Given the equation of a circle and a chord length, how can you find the shortest distance from the center to the chord?

-First, find the radius of the circle from its equation. Then, knowing half the chord length, use the Pythagorean theorem to calculate the shortest distance, which is the perpendicular distance from the center to the chord.

### Why is it important to understand the relationship between the radius, tangent, and chord in coordinate geometry problems involving circles?

-Understanding these relationships is crucial for solving problems involving the equation of tangents, normals, and finding distances within the circle, which are common in coordinate geometry.

### Outlines

### π Introduction to Circles in Coordinate Geometry

This paragraph introduces the topic of circles within the context of Year 12 coordinate geometry, following a previous discussion on lines. The focus is on understanding the equation of a circle, both in standard form and when it requires completing the square to reveal the center and radius. An example is given where the equation 'x squared plus 14x plus y minus 6 squared equals 2 squared' is simplified to find the circle's center at (-7, 6) and a radius of 5. The importance of recognizing that the radius is the square root of the constant term in the equation is emphasized, as is the concept that the radius is perpendicular to the tangent of the circle.

### π Applications of Circle Geometry: Tangents and Perpendicular Bisectors

The second paragraph delves into the applications of circle geometry, specifically discussing the relationship between the radius and the tangent, and the significance of the perpendicular bisector of a chord. It explains how to derive the equation of a tangent line using the gradient of the radius and the point of tangency, and how the perpendicular bisector of a chord passes through the center of the circle. A problem is presented where the length of a chord is given, and the task is to find the shortest distance from the center of the circle to the chord. The solution involves using the perpendicular bisector, recognizing right angle triangles formed, and applying the Pythagorean theorem to find this shortest distance, which is demonstrated with the given circle equation and chord length.

### Mindmap

### Keywords

### π‘Circle

### π‘Equation of a Circle

### π‘Complete the Square

### π‘Center of a Circle

### π‘Radius

### π‘Tangent

### π‘Gradient

### π‘Perpendicular Bisector

### π‘Chord

### π‘Pythagorean Theorem

### Highlights

Introduction to the topic of circles in coordinate geometry within the A-Level Maths curriculum.

Explanation of how to derive the equation of a circle from its standard form, including identifying the center and radius.

Demonstration of completing the square to find the center and radius of a circle when not in standard form.

Example problem to illustrate the process of completing the square for a circle equation.

Clarification on the importance of correctly identifying the center coordinates, especially with negative signs.

Mistake to avoid: confusing the value inside the equation with the actual radius.

The concept that the radius is perpendicular to the tangent of a circle.

Application of the perpendicular radius to tangent concept in finding the equation of a tangent line.

Method to determine the gradient of a tangent line using the perpendicular property to the radius.

Using the point-slope form to write the equation of a tangent line to a circle.

The geometric property that the perpendicular bisector of a chord passes through the center of a circle.

Practical application of the perpendicular bisector in finding the shortest distance from the center to a chord.

Utilization of right-angle triangles and Pythagoras' theorem to solve for the shortest distance.

Step-by-step guide on how to find the radius of a circle given its equation.

Final solution to the example problem using the derived radius and Pythagoras' theorem.

Conclusion summarizing the key points covered in the revision on circles within coordinate geometry.

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