|
|
Revision:OCR Core 1 - Circles
From The Student RoomTSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > OCR Core 1 - Circles
13. CirclesThe equation of a circleOne must be able to describe the circle as an equation on the Cartesian plane. Consider an application of Pythagoras' Theorem. Construct a right-angled triangle whose corner touches the Origin, and whose right-angle allows one of the shorter sides to form a perpendicular with the x-axis. Also, one says that the hypotenuse connects the Origin to a point (above the point at which the right-angle occurs, on the x-axis). If one is to allow the hypotenuse to be constant, and known, one can calculate the vertical side of the triangle (which is perpendicular to the x-axis) as one varies "x" (the side of the triangle which is on the x-axis). As one allows "x" to increase and decrease, it forms a circle whose radius is the hypotenuse. Hence one can give the equation thus:
This is merely Pythagoras' Theorem, however it produces the following (note that the constructed triangle is a demonstration of the text above, it is not part of the graphed function). Figure 007 This shows how the circle can be centred at the Origin. Now suppose that the centre is at the point One can suppose also that there is a point upon the circumference, This means that one can apply the first technique that was discussed in these notes, regarding the distance between two points. The radius is therefore equal to:
Hence, one can now recreate a more general equation for the circle (on the Cartesian plane):
Where This can now be applied in a number of ways. Often one will see the formula in different forms, however, if one wishes to obtain the centre, or radius figures quickly, it is better to refactor the equation such that it is in the aforementioned form. To sketch a circle from an equation, one must simply get the given equation into the aforementioned form (often requiring the completion of the square technique), and then one can obtain the details necessary (namely the centre, and the size of the radius). Some background knowledgeOne must be aware of the circle theorems in order to do well at questions involving the equation of a circle. One of the most required theorems to know is that if there is a tangent to a circle, the diameter at the point of contact will be at right-angles to the tangent (this is useful in calculations regarding the normal). Another useful circle theorem is that if there is a chord, and a diameter cuts it in a perpendicular manner, that diameter is a perpendicular bisector of the chord (in other terms, the chord is intersected half way along it's length by the diameter). Another useful property is that if one is to take a diameter, and then take a point upon the circumference, and join the diameter to the point on the circumference (forming a triangle), the angle at the circumference is always a right angle. These notes do not prove these assertions, one might take it upon oneself to look into the proofs of these, and other theorems regarding the circle. Lines and circlesAs one is unable to differentiate the equation for a circle at the current time (assuming that one knows only the contents of this module), one must use a different manner of calculating the gradient (and hence the equation of) the tangent to a circle at a given point. The technique is simple. As one is aware of the centre (after refactoring the equation), and radius of the circle, one can use the coordinates from the centre of the circle along with those of the point upon the circumference to obtain a gradient for a line (the normal) which is perpendicular to the tangent. Then one can use the rule discussed previously to obtain the gradient of the tangent, and finally one can apply these methods with the method involving the general equation for a straight line in order to obtain the equation for the tangent (or the normal for that matter).
Also SeeRead these other OCR Core 1 notes:
Comments |











.
.
, and
.





