Coordinate geometry of the circle

Question 1

How can a circle with centre (a,b) and radius r be written in completed square form?

Answer 1

(x-a)²+(y-b)²=r²

Question 2

What is general form?

Answer 2

x²+y²+2gx+2fy+c=0 where (-g,-f) is the centre and root(g²+f²-c) is the radius

Question 3

Intersecting circles

Answer 3

AB < r₁+r₂

Question 4

Touching circles

Answer 4

AB = r₁+r₂

Question 5

Touching circles

Answer 5

AB = r₁-r₂

Question 6

Non-intersecting circles

Answer 6

AB > r₁+r₂

Question 7

Non-intersecting circles

Answer 7

AB < r₁-r₂

Question 8

Determine whether (x-5)²+(y-2)² = 13 and y = x-4 intersect

Answer 8

1. Substitute: (x-5)²+((x-4)-2)² = 13
2. Rearrange and simplify: x²-11x+24 = 0
3. Use discriminant: If b²-4ac > 0, 2 points of intersection, chord. If b²-4ac = 0, 1 points of intersection, tangent. If b²-4ac < 0, 0 points of intersection

Question 9

Find the equation of a circle whose centre is (2,-1) and has tangent y+3x = 0

Answer 9

Gradient of tangent is -3
Gradient of radius is 1/3
(y-(-1))/(x-2) = 1/3
(y+1)/(x-2) = 1/3
3(y+1) = 1(x-2)
3y+3 = x-2
3y = x-5
3y = -9x
-9x = x-5
-10x = -5
x = 1/2
y = -3/2
Point of intersection in (1/2,-3/2)
root((2-1/2)²+(-1–3/2)²) = root(10/4)
(x-2)²+(y+1)² = 10/4