The radical centre of the circle x2+y2+2x+6y=0, x2+y2-4x-2y+6=0 and x2+y2-12x+2y+30=0 is
Answers
Given: Three equations: x^2+y^2+2x+6y=0, x^2+y^2-4x-2y+6=0 and x^2+y^2-12x+2y+30=0
To find: The radical centre of the circle.
Solution:
- Now we have provided three equations, let them be: S1, S2 and S3.
S1 = x^2 + y^2 + 2x + 6y = 0
S2 = x^2 + y^2 - 4x - 2y + 6 = 0
S3 = x^2 + y^2 - 12x + 2y + 30 = 0
- Now to find the radical axis, we will do:
- S2 - S3 = 0
x^2 + y^2 - 4x - 2y + 6 - ( x^2 + y^2 - 12x + 2y + 30 ) = 0
x^2 + y^2 - 4x - 2y + 6 - x^2 - y^2 + 12x - 2y - 30 = 0
8x - 4y - 24 = 0
8x - 4y = 24 ..........................(i)
- S3 - S1 = 0
x^2 + y^2 - 12x + 2y + 30 - ( x^2 + y^2 + 2x + 6y ) = 0
x^2 + y^2 - 12x + 2y + 30 - x^2 - y^2 - 2x - 6y = 0
-14x - 4y + 30 = 0
14x + 4y = 30.........................(ii)
- S1 - S2 = 0
x^2 + y^2 + 2x + 6y - ( x^2 + y^2 - 4x - 2y + 6 ) = 0
x^2 + y^2 + 2x + 6y - x^2 - y^2 + 4x + 2y - 6 = 0
6x + 8y - 6 = 0
6x + 8y = 6 ........................(iii)
- From (i) and (ii), we have:
8x - 4y + 14x + 4y = 24 + 30
22x = 54
x = 2.45
- Putting x in (i), we get:
8(2.45) - 4y = 24
y = -1.1
Answer:
So (2.45,-1.1) is the radical centre.