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Another Concentric Circle Question?
Need to find the equation to a concentric circle.
x^2 + y^2 = 6x passes through (4, -3)
The answer is x^2 - 6x + y^2 -1 = 0
I do get the above answer if I first write it out as x^2 + y^2 = 6x + 4 - 3 then move everything to the other side but I do have quite the feeling that's the incorrect way to do it.
Thanks to both of you, I sort of see where to go but still somewhat confused while trying to compare this to what is in my text book.
2 Answers
- Mike GLv 78 years agoFavorite Answer
Complete the square for circle given
x^2-6x+9+y^2 = 9
(x-3)^2+y^2 = 3^2
Circle has centre (3,0)
Our concentric circle also has centre (3,0)
Equation is
(x-3)^2+y^2 = r^2 and has a point (4,-3)
1^2 + (-3)^2 = r^2
r^2 = 10
Equation is
(x-3)^2+y^2 = 10
x^2-6x+9+y^2 = 10
x^2-6x+y^2-1 = 0
- PeterLv 48 years ago
x^2 + y^2 = 6x does NOT pass through (4,3). The LHS is 16+9 = 25, the RHS is 6*4 = 24.
The question most is most likely (though not definitely)
x^2 + y^2 + c = 6x
If you substitute in (4, -3) you get c = -1
So x^2 + y^2 -1 = 6x
x^2 + y^2 - 6x - 1 = 0
Which is pretty much what you have done, but without explaining (or possible even understanding) why you have done it.
BTW, this expression can be manipulated as follows:
x^ - 6x + 9 - 10 + y^2 = 0
(x-3)^2 + y^ = 10
Which better shows (IMHO) what the circle looks like - it has radius of sqrt(10) and is centred at (3,0).