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P2.3 - Coordinate geometry in the (x, y) plane

Syllabus
2019
Topic
P2.3
Level
AS

Learning objectives

P2.3.1Coordinate geometry of the circleCoordinate geometry of the circle Students should be able to find the radius and the using the equation of a circle in the coordinates of the centre of the circle, given the equation of form (x − a)2 + (y − b)2 = r2 and the circle and vice versa. including use of the following circle properties: (i) the angle in a semicircle is a right angle.; (ii) the perpendicular from the centre to a chord bisects the chord.; (iii) the perpendicularity of radius and tangent.

Read a circle equation and unlock its geometry

A circle is the set of points a fixed distance rr from its centre (a,b)(a,b). Its coordinate equation records that distance using Pythagoras:

(x−a)2+(y−b)2=r2(x-a)^2+(y-b)^2=r^2

Read the centre as (a,b)(a,b) and the radius as the positive square root of the right-hand side. The signs inside the brackets are reversed. To build an equation, substitute the known centre and radius; if a diameter's endpoints are known, its midpoint is the centre and half its length is the radius.

For x2+y2−6x+4y−12=0x^2+y^2-6x+4y-12=0, move the constant and add 99 and 44 to both sides: x2−6x+9+y2+4y+4=12+9+4,x^2-6x+9+y^2+4y+4=12+9+4, so (x−3)2+(y+2)2=25.(x-3)^2+(y+2)^2=25. Thus the centre is (3,−2)(3,-2) and the radius is 55.

Circle property Coordinate-geometry use
If ABAB is a diameter and CC is on the circle, ∠ACB=90∘\angle ACB=90^\circ prove or use a right angle
A perpendicular from the centre to a chord bisects the chord locate a chord midpoint or centre
The radius OPOP is perpendicular to the tangent at PP find a tangent or normal gradient

In the example, P=(7,1)P=(7,1) lies on the circle because 42+32=254^2+3^2=25. The radius CPCP has gradient 3/43/4, so the tangent gradient is −4/3-4/3 and its equation is y−1=−43(x−7)y-1=-\frac43(x-7).

Do not read rr directly from r2r^2, or copy the bracket signs into the centre. The radius-tangent rule applies at the point of contact; perpendicular non-vertical gradients satisfy m1m2=−1m_1m_2=-1.