Wilson’s School Core 1 Revision Sheet 3 of 5
Coordinate geometry
Stuff to memorise:
Straight lines with gradients m1 and m2 are perpendicular when m1m2  1
The equation of a circle with centre (a, b) and radius r is
( x  a) 2  ( y  b) 2  r 2
I can:
Find the length, gradient and mid-point of a line-segment
Find the equation of a straight line given sufficient information
Understand the equation of a circle and solve problems involving
lines and circles
Test yourself:
1 Find the centre and radius of the circle with equation
x 2  y 2  10 x  14 y  10  0
2 The points (10,-4) and (2,2) are the ends of the diameter of a
circle. Find the equation of the circle.
3 Find the equation of the line joining the points (-2,4) and (-4,8)
4 Find the perpendicular distance from (2,4) to the line y=2x+10.
5 Find the equations of the tangent and normal to the circle
x 2  y 2  6 x  10 y  2  0 at the point (-2, 3).
Tick!
Answers:
1
Complete the square to give
( x  5) 2  25  ( y  7) 2  49  10  0
( x  5) 2  ( y  7) 2  64
Hence centre (-5,7) radius 8
2
 2  10 2  4 
,
Centre is the midpoint 
  (6,1)
2 
 2
Radius is length of line from centre to one of the points
r
2  62  2  12
 25  5
Hence equation of circle is x  6   y  1  25
84
4
3

 2
Gradient of line is
 4  2  2
y  4  2 ( x  2 )
Using y  y1  m( x  x1 ) gives
y  2x  0
4
2
2
1
2
y  4   1 2 ( x  2)
Using y  y1  m( x  x1 ) gives equation of the perp line
y   12 x  5
These two lines intersect when
Gradient of line is 2 hence gradient of perpendicular line is
 12 x  5  2 x  10
 x  10  4 x  20
x  2, y  6
Hence the distance required is the distance from (2,4) to (-2,6)
2  22  4  62
x  32   y  52  36
d
5
 20  2 5
Centre (-3,-5) radius 6
Gradient from centre to point [ ie of radius] =
Equation of normal, using y  y1  m( x  x1 )
y  3  8( x  2)
0  8 x  y  19
Gradient of tangent =
1
8
Equation of tangent, using y  y1  m( x  x1 )
1
y 3 
( x  2)
8
8 y  24   x  2
x  8 y  22  0
3  5
8
 2  3