Origin-centred circles and tangent equations · Higher
| English | 中文 | Pinyin |
|---|---|---|
| radius gradient/ˈreɪdɪəs ˈɡreɪdɪənt/ | 半径斜率 | bàn jìng xié lǜ |
Which line just touches the circular path?
- A circular path has a sensor at (3,4). Which straight line just touches the path at that point?
- This lesson studies radius gradient 半径斜率: The slope of a radius joining the centre to a point on the circle.
Choose the mathematical structure
- For an origin-centred circle, x²+y²=r². A point is on the circle if its squared coordinates sum to r². A tangent is perpendicular to the radius there. Use a negative reciprocal gradient when both gradients are finite; handle horizontal/vertical cases directly.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines radius gradient?
The slope of a radius joining the centre to a point on the circle.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
At (3,4), r²=3²+4²=25, so x²+y²=25. The radius gradient is 4/3, hence tangent gradient -3/4. Its equation y-4=(-3/4)(x-3) simplifies to 3x+4y=25. At (5,0) the radius is horizontal and the tangent is the vertical line x=5.
Origin-centred circles and tangent equations
For an origin-centred circle, x²+y²=r²
Compare the model with the worked case and explain one change.
Find the radius of the circle through (3,4).
√(3²+4²)=5.
Test a tempting shortcut
- The radius and tangent share a point but different directions. Do not use the negative reciprocal of zero; a horizontal radius has a vertical tangent. Check the tangent passes through its contact point.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
The tangent at a circle point has the same gradient as the radius to that point. This claim is false. Explain which definition or assumption it violates.
Find the tangent gradient at (3,4).
The negative reciprocal of 4/3 is -3/4.
The tangent at a circle point has the same gradient as the radius to that point.
The radius and tangent share a point but different directions. Do not use the negative reciprocal of zero; a horizontal radius has a vertical tangent. Check the tangent passes through its contact point.
Interpret a new situation
- AQA A16 Higher concerns circles centred at the origin. Offset-centre circle formulae and circle calculus are not needed for this lesson.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
For tangent 3x+4y=25, find x when y=0.
Set y=0, then divide 25 by 3.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 8300 · Higher · 3.2. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The slope of a radius joining the centre to a point on the circle. Choose the relationship, show the method, check its assumptions and interpret the result.