answers:
(x+3)^2+(y+2)^2=10
(x-3)^2+(y+2)^2=10
(x-2)^2+(y+3)^2=10
(x+2)^2+(y-3)^2=10
2007-03-21 03:48:13 · 5 answers · asked by tia b 1 in Science & Mathematics ➔ Mathematics
Colliate all x and y terms
x^2 - 6x + y^2 + 4y + 3.
Next we have to examine the options available
(x+3)^2 would equal x^2 + 6x + 9 , so this is wrong
(x-3)^2 would equal x^2 -6x + 9, so this look right.
(x-2)^2 would equel x^2 - 4x +4, so this is wrong
(x+2)^2 would equal x^2 +4x + 4 s o this is wrong
so then (x-3)^2+(y+2)^2=10 is correct
2007-03-21 03:56:38 · answer #1 · answered by The exclamation mark 6 · 0⤊ 0⤋
standard form of a circle is
(x - a)^2 + (y - b)^2 = r^2
your equation is... x^2 - 6x + y^2 + 4y + 3 = 0
take the coefficient of x and y... divide both by 2... and then square the terms... add and substract those terms from your equation...
coefficient of x is -6... -6/2 is -3... square of -3 is 9... add and subtract 9...
coefficient of y is 4... 4/2 is 2... square of 2 is 4... add and subtract 4...
now you have x^2 - 6x + 9 - 9 + y^2 + 4y + 4 - 4 + 3 = 0...
rearrange the equation...
x^2 -6x + 9 + y^2 + 4y + 4 + 3 - 9 - 4 = 0
find the factors of 9 such that if you add them, you have -6...
-3*-3 = 9 and -3 + (-3) = -6
find the factors of 4 such that if you add them, you have +4...
2*2 = 4 and 2+2 = 4...
therefore, your equation now reads...
(x - 3)^2 + (y + 2)^2 - 10 = 0
(x - 3)^2 + (y + 2)^2 = 10
so answer is number 2
2007-03-21 04:03:08 · answer #2 · answered by Faraz S 3 · 0⤊ 0⤋
first question: gradient of line = -2.5 y-intercept = 5 therefore equation of line is y= -2.5x + 5 2d question: 6x - 4y = 12 may be decreased to 4y = 6x - 12 and then y = a million.5x - 3 therefore its gradient is a million.5, so the gradient of the perpendicular line ought to ought to -2/3 2x + ky = 6 may be decreased to ky = -2x + 6 and then y = (-2/ok)x + (6/ok) -2/ok = -2/3 therefpre ok = 3
2016-12-02 08:39:12 · answer #3 · answered by allateef 4 · 0⤊ 0⤋
x^2+y^2-6x+4y+3=0
x^2-6x+y^2+4y+3=0
x^2-2*x*3+3^2+y^2+2*y*2+2^2+3-3^2-2^2=0
(x^2-2*x*3+3^2)+(y^2+2*y*2+2^2)+(3-3^2-2^2)=0
(x-3)^2+(y+2)^2+3-9-4=0
(x-3)^2+(y+2)^2-10=0
(x-3)^2+(y+2)^2=10
[ (a+b)^2=a^2+2ab+b^2 and (a-b)^2=a^2-2ab+b^2]
2007-03-21 04:03:20 · answer #4 · answered by Bubblez 3 · 0⤊ 0⤋
Just by inspecting the options one can point out at option 2.
Try to develop techniques that save time
2007-03-21 03:56:09 · answer #5 · answered by Keeper of Barad'dur 2 · 0⤊ 0⤋