Everything Maths Grade 12

9.3 CHAPTER 9. GEOMETRY

9.3 Co-ordinate Geometry

Equation of a Circle EMCBZ

We know that every point on the circumferenceof a circle is the same distance away from the centre of

the circle. Consider a point (x 1 ;y 1 ) on the circumference of a circle of radius r with centre at (x 0 ;y 0 ).

�

�

(x 0 ;y 0 )

P (x 1 ;y 1 )

O Q

Figure 9.3: Circle with centre (x 0 ;y 0 ) and a point P at (x 1 ;y 1 )

In Figure 9.3,�OPQ is a right-angled triangle. Therefore, from the Theorem of Pythagoras, we know

that:

OP^2 = PQ^2 +OQ^2

But,

PQ = y 1 −y 0

OQ = x 1 −x 0

OP = r

∴ r^2 = (y 1 −y 0 )^2 + (x 1 −x 0 )^2

But, this same relation holds for any point P on the circumference.Therefore, we can write:

(x−x 0 )^2 + (y−y 0 )^2 = r^2 (9.3)

for a circle with centre at (x 0 ;y 0 ) and radius r.

For example, the equation of a circle with centre (0; 0) and radius 4 is:

(y−y 0 )^2 + (x−x 0 )^2 = r^2

(y− 0)^2 + (x− 0)^2 = 4^2

x^2 +y^2 = 16

See video: VMhrm at http://www.everythingmaths.co.za