Expert: Josh - 1/27/2010

Question
1.Determine the equation of the circle through the points (3;2)(5;-4)with the centre on the line x-y=1

2.determine the equation of a circle with diameter ab where a is the point (-3;2) and b is the point (1;6)

Answer
Dear Zainy,

A circle with center (x=a,y=b) and radius "r" can be described by the equation (x-a)^2+(y-b)^2=r^2 .....[#1]
Obviously, our goal is to determine the values of "a", "b" and "r".

If the center is constrained to lie somewhere on a straight line x-y=1, it must satisfy the condition a=1+b ...[**] if we make a substitution (x=a, y=b) for the unknown center coordinates.

Using this information, [#1] becomes (x-1-b)^2 + (y-b)^2 = r^2 .....call this equation [#2].

Now, we were told the circle passes through (x=3,y=2) and (x=5,y=-4). So, these points must satisfy the equation in [#2]. This means:

Substituting (x=3,x=2),
(3-1-b)^2 + (2-b)^2 = r^2
(2-b)^2 + (2-b)^2 = r^2
2[4-4b+b^2] = r^2
8 -8b +2b^2 = r^2   .....[#3]

Substituting (x=5,x=-4),
(5-1-b)^2 + (-4-b)^2 = r^2
(16-8b+b^2) + (16+8b+b^2) = r^2
32+2b^2 = r^2    .....[#4]

[#3] and [#4] leave us with 2 equations with 2 unknowns (in terms of "b" and "r").

To proceed, we may, for instance, substitute [#4] viz., r^2 = 2b^2+32 into the RHS of [#3].

This gives 8 -8b +2b^2 = 2b^2 +32. The "b square" terms cancel on both sides. 8-8b=32 remains. Simple rearrangement gives -24=8b. At long last, b=-3.

Using back-substitution, putting b=-3 into [#4], we get r=sqrt(50) after simplification. (Note: when we take the square root, we also get "-sqrt(50)". But this is not an acceptable solution. We reject it because "r" represents the radius and must be non-negative).

From the condition (**), a=1+b, so a=-2.

Altogether, a=-2, b=-3, r=sqrt(50) in (x-a)^2 + (y-b)^2 = r^2.
I tend to leave it in this form since it is more intuitive.

Q2. Work out the diameter (d) using Euclidean distance.
d = sqrt[(1--3)^2 + (6-2)^2] = sqrt(32) = 4*sqrt(2).
Hence, radius r = d/2 = 2*sqrt(2).

Again, you can use a similar approach to Q1.
Start with (x-a)^2 + (y-b)^2 = r^2  ....[#5]
This time we already know r=2*sqrt(2).
Form two equations by substituting the points (x=-3,y=2) and (x=1,y=6) into [#5].
You will have two equations with two unknowns (in "a" and "b"). This, you can solve for.
I won't go through the steps here, as the procedure was explained in Q1.