- Edexcel, Module - C2, Chapter - Coordinate Geometry
- AQA, Module - C1, Chapter - Coordinate Geometry
- OCR, Module - C1, Chapter - Coordinate Geometry

In C1, part of the Coordinate Geometry Chapter, we learnt about linear equations and graphing them. In C2, we learn about the equations of circles, from which we can graph them, aswell as solving problems again involving tangents. Remember in GCSE, we learn about

**Circle Theorems**, here are a few we need to know, which will come in use in this chapter :

**Circle Terminology !****- The distance around a circle, given by the formula**

*Circumference***2pi * r.**

**- The distance from one point on the circumference, to another, passing through the centre of the circle. It is twice the radius (2r)**

*Diameter***- A line which is from the centre of the circle, to any point of on the circumference of the circle. It is half the diameter.**

*Radius***Area of a circle is given by the formula - pi * (r^2)**

**Circle Theorems Recap !**1) The

**angle in a semicircle is a right angle**(An angle formed by drawing lines from the ends of a diameter to its circumference, will form a right angle)

2) The

**perpendicular from the centre of the chord, bisects the chord**.

(A

**chord**is a line that has both endpoints on the circumference on the circle.)3)The

**perpendicularity of the radius and tangent.**

(A tangent to the circle, forms a right angle, with the circle's radius, at the point of contact with a tagnent).

Showing the 3rd theorem, which shows a line from the centre of the circle till the circumference (Radius), when it meets with a tangent to the circle at a point, it will form a right angle ! |

**Equation of a Circle**The equation of a circle is given in the form :

**(x-a)**

^{2}+ (y-b)^{2}= r^{2 }

^{ }

^{a = x -coordinate of centre}

^{b= y-coordinate of centre}

^{r= radius}^{e.g.1 ) What is the equation of a circle when the centre of the circle is (-4,5) and the diameter is 10.}

^{ }

^{centre of circle (-4,5), so a = -4 and b = 5}

^{radius = half the diameter so 10/2 = 5}

^{so }(x + 4)

^{2}+ (y - 5)

^{2}= 5

^{2 }

^{ }

^{Just leave it in the form 5 squared, rather than writing 25.}

^{e.g.2) What is the centre of the circle, and radius, given this equation of the circle :}

^{ }

(x)

^{2}+ (y+8)

^{2 }= 49

^{ }

^{Centre of Circle -a= 0, and b = 8. So centre of circle is (0, -8 )}

^{Radius is the square root of 49, some people still write + / - 7, even knowing a length can't be negative. so the radius is 7.}

^{ }

^{We've met our first point, knowing how to derive the equation of a circle, given the centre points and the radius, and vice versa.}

^{Finding the centre of a circle, through Completing the Square }

^{ }

^{If we are given the equation of the circle to be :}

*x*

^{2}+ 6

*x*+

*y*

^{2}- 8

*y*- 11 = 0 , and we have to determine the centre of the circle and radius. Here, we can see it's not in the usual form, but still it is an equation of a circle. We just have to complete the square (look under the c1 completing the square post if you've forgotten) :

^{complete the square for x terms and y terms seperately:}

^{so }

*x*

^{2}+ 6

*x becomes =*

*(x+3)*

^{2}- 9

so

*y*

^{2}- 8

*y becomes =*

*(y - 4)*

^{2}- 16

we move the -11 to the other side, so we get :

*(x+3)*

^{2}- 9 +

*(y - 4)*

^{2}- 16 = 11

*(x+3)*

^{2}+

*(y - 4)*

^{2}- 25 = 11

*(x+3)*

^{2}- 9 +

*(y - 4)*

^{2}= 36

*(x+3)*^{2}+*(y - 4)*^{2}=*6*^{2}^{Now, we have to the equation of the cirlce in the desired form :}

^{ }

*(x- a)*^{2}+*(y - b)*^{2}=*r*^{2}^{The }

^{centre of the circle is (-3,4) and the radius is 6.}^{This is all to the coordinate geometry chapter, in terms of what we need to know. In the next post, I will post some questions and their model solutions that are quite common, which require some knowledge of GCSE theorems!}

^{ }

^{ }
I ma here to discuss about circle as circle making is a easy task just draw a curve in response to a fix line from a central point and all points are the the same distance from the center.

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