Find a for which the points (1,2), (-3,2),(6,5) and (a,3) are concyclic.
I used the properties but I had a very heavy polynomial which broke my heart
shouldn't be too heavy, since you're just dealing with a circle through 3 points.
If we call the first three points A,B,C, then
AB: midpoint = (-1,2), slope = 0, perpendicular bisector is x = -1
BC: midpoint = (3/2,7/2), slope = 1/3, pb is y = x/3 + 3
The pb's intersect at (-1,8/3)
So r^2 = 2^2 + (2/3)^2 = 40/9
and thus the circle is (x+1)^2 + (y - 8/3)^2 = 40/9
So, (a+1)^2 + (3 - 8/3)^2 = 40/9
(a+1)^2 + 1/9 = 40/9
a = -1 ±√39/3
I suggest you find the equation of the circle using the first 3 points.
I noticed that (1,2) and (-3,2) would yield a horizontal chord, and therefore
the right-bisector would be x = -1, and the centre must have x = -1
then find the equation for the right-bisector of the chord from (1,2) to (6,5)
the centre will be the intersection of x = -1 and the right-bisector equation
From there it is easy to find the equation, you should ge
(x+1)^2 + (y-1)^2 = 85
sub in the point (a,3) to find a
Thank you so so much reiny and obleck
one or both of us is wrong, though, since we disagree. Better check our math.
I was wondering
I got a=2√2 for reiny
I intend using this properties
AC×BD=(AD×BC)+(AB×CD)
hoping to get some expression to find a but it became so tedious I left it
I'll check the maths
Sorry got 8
The equation of the circle is (x+1)^2 + (y-1)^2 = 85
(All 3 points satisfy the equation)
so for (a,3)
(a+1)^2 + 4 = 85
(a+1)^2 = 81
a+1 = ± 9
a = -10 , or a = 8
I'm not sure how you justify your equation. Since A,B,C,D all lie in the same plane, (AD×BC) and (AB×CD) are both vertical vectors, and adding them will not give you any useful information. Or is there some theorem about cross products and concyclic points?
No I had a wrong thinking then thank you but i read some on it and the said something on complex number using argument
Arg[(z4-z1)/(z2-x1)]+arg[(z2-z3)/(z4-z3)=π
When they multiplied it they said it was real