O Level A Maths Eqn of Circle Qn

• quailmaster

  11 Aug 13, 07:34

  FInd the equation of circle which passes through the points A(0,1), B(3,-2)  and has its centre lying on the line y = x-2

  Ans given is x^2 + y^2 - 2x + 2y -3 = 0

• quailmaster

  11 Aug 13, 10:09

  I tried finding perpendicular bisector but the eqn of bisector same as eqn given in qn

• FireIce

  11 Aug 13, 12:55

  let O be x,y

  length OA = length OB (radius)

  simultaneous eqn

• quailmaster

  11 Aug 13, 13:15

  Hi I tried that as well, u end up with eqn similar to the one given

   

• FireIce

  11 Aug 13, 13:49

  

   

  expand

  arrange to one side

• Seowlah

  17 Aug 13, 00:23

  Wow, don't play, play hor

  FireIce also answered Maths questions 

• dreamer87

  17 Aug 13, 02:40

  There is not enough information to solve this equation. 

  You need at least 3 pts to determine the circle uniquely. 

  P.S That what I think. 

• Seowlah

  30 Sep 13, 13:26

  The answer given by TS is incorrect, although by substituting the coordinates of A and B do satisfy the equation. Based on the incorrect answer given the centre is (1, -1). But when the plotting of the points of A and B, centre (1, -1) and the line y = x - 2 are done on a graph paper, a circle cannot be drawn to fulfil all the conditions. Hence, the given answer is incorrect.

  Steps

  (1)  The usual way to find the centre in this type of question is to find the perpendicular bisector of AB and find the intersection point of the perpendicular bisector of AB and the line y = x - 2. However, in this question, the perpendicular bisector of AB is the same as the line y = x - 2 where the centre lies and the centre of the circle cannot be found in this way.

  However, it is to be noted the perpendicular bisector of AB and the line y = x - 2 are the same means that the perpendiculr bisector is the diameter of the circle.

  (2)  Hence, to find the centre, we just need to find the mid-point of AB =(1.5, - 0.5).

  (3)  Next, to find the radius, we will use point A(0,1) and Centre (1.5, - 0.5) to get the radius of square root of 4.5

  (4)  Thus the equation of the circle is (x - 1.5)^2 + (y + 0.5)^2 = 4.5

  To check, substituting the coordinates of A and B into the equation of the circle, it is found to satisfy the equation. A circle can also be drawn to fulfil all the conditions when these conditions are plotted on a graph paper.

   

   

• Seowlah

  30 Sep 13, 13:41

  Question

  FInd the equation of circle which passes through the points A(0,1), B(3,-2)  and has its centre

  lying on the line y = x - 2

  Solution

  Mid-point of AB = [ (0+3) /2 , (1 + ( -2)) ] = (1.5, - 0.5)

  Gradient of AB = ( -2 - 1) / (3 - 0) = - 1

  Gradient of Perpendicular bisector of AB = - 1 / - 1 = 1

  Subsituting midpoint (1.5, - 0.5 ) and gradient of 1 into

  y = mx + c

  - 0.5 = 1 (1.5) +c

  c = - 2

  So, the equation of the perpendicular bisector of AB is 

  y = mx + c

  y = 1x - 2

  Hence, the equation of the perpendicular bisector of AB is the SAME as the line

  y = x - 2 where the centre of the circle lies.

  THUS, the equation of the perpendicular bisector of AB is the diameter of the circle

  and the midpoint (1.5, - 0.5) is also the centre of the circle.

  Using A(0,1) and midpoint (1.5, - 0.5), 

  radius  = square root [(1.5 - 0)^2 + (- 0.5 - 1)^2]

  radius = square root (4.5)

  Hence, the equation of the circle is (x - 1.5)^2 + (y + 0.5)^2 = 4.5