The Circle C has the equation x^2 + y^2 - 12x + 8y +16 = 0?

a) Find the coordinates of the centre of C


b) Find the radius of C


c) Sketch C


Given that C crosses the x-axis at the points A and B


d) Find the length AB, giving your answer in the form K√5





Its the last question i'm stuck on for my homework, any help would be great :)

The Circle C has the equation x^2 + y^2 - 12x + 8y +16 = 0?
a) General equation of a circle can be expressed in


x^2 +y^2 +2gx+2fy +c = 0 with (-g,-f) as centre and


SQRT(g^2 + f^2 -c)


=%26gt; Hence the above equation can be transformed to


x^2 + y^2 + 2(-6)x + 2(4)y + 16 = 0 ,


i.e g = -6 , f = 4 , hence the centre of the above circle is (6,-4) .





b) Radius = SQRT(6^2 + (-4)^2 - 16)) = SQRT(36) = 6 Units





c) You can draw the Circle with (6,-4) as centre and 6 as radius on a coordinate axes.





d) For solving this , put y = 0 in the equation of the circle and solve for point A %26amp; B.





=%26gt; x^2 + 0^2 - 12x + 8x0 + 16 = 0


=%26gt; x^2 -12x + 16 = 0


Solving for x , you will get the coordinates for A and B as follows:A(6+2√5,0) and B(6-2√5,0). Hence the length of AB


= 4√5 units
Reply:We need to put this in the form (x-a)^2 + (b-y)^2 = r^2, and we get (x-6)^2 -36 + (y+4)^2 -16 +16 = 0. This simplifies to (x-6)^2 + (y+4)^2 = 36. This takes care of parts a and b. Substitute y = 0 in this, solve for the two values of x, take their difference, and you're done.
Reply:Perhaps this may help:





The center of a circle is defined as (h,k)





h is the x value, k is the y value





in order to find (h,k) you have to get the equation into the form





(x-h)^2 + (y-k)^2 = r^2 r here is the radius





to put your equation in to the right form you must group the x values and the y values.





so you get: x^2 -12x + y^2 + 8y= - 16


(note that the 16 was brought over to the other side)





now lets work with x and y separately





right now we have x^2 - 12X


we must complete the square. take the -12 and cut it in half.


we now have 6. square this number to get 36





[Make sure you add the 36 to the -16 on the other side of the equal sign!]





so now you have x^2-12x+36


Factor this to get (x-6)^2





now do the same for the y: cut the 8y in half, square it, add that value to the -16 on the other side, and factor.


this should lead you to (y+4)^2





now all in all you should have (x-6)^2 + (y+4)^2 = 36


(the 36 comes from [ -16+36+16]





so looking at this equation we can tell that the center of the circle is (6, -4)


to find the radius we remember that (x-h)^2+(y-k)^2=r^2


in this case r^2 = 36 so r equals the square root of 36 which equals six.


I hope this helps you with the first questions of your homework and should you need further help you might find it at http://www.analyzemath.com/CircleEq/Tuto...
Reply:(x-6)^2 + (y+4)^2 = 36+16-16 = 36





[(x-6)/6]^2 + [(y+4)/6]^2 = 1





origin (6,-4)


radious 6
Reply:center is(-a/2,-b/2)=(12/2,-8/2)=(6,-4)
Reply:As many have said,


(x - 6)^2 + (y + 4)^2 = 36.





The x-axis is y=0, so the two x-values are the solutions of


(x - 6)^2 + (0 + 4)^2 = 36;


i.e.


(x-6)^2 = 20; x = 6 ± √20 = 6 ± 2√5.





The distance between the two is


(6 + 2√5) - (6 - 2√5) = 4√5
Reply:First of all, complete the square to get the equation in the form


(x - x0)^2 + (y - y0)^2 = r^2





(x^2 - 12x) + (y^2 + 8y) = -16





(x^2 - 12x + 36) + (y^2 + 8y + 16) = -16 + 36 + 16





(x - 6)^2 + (y + 4)^2 = 6^2





a) From the equation above, x0 = 6 and y0 = -4, so the coordinates of the center are (6, -4).





b) Again, from the equation above, r = 6.





c) Simply draw a circle centered at the point (6, -4) with a radius of 6 units.





d) I'm not sure what you mean by AB. More information is needed.
Reply:x^2 - 12x + y^2 + 8y = -16.


(x - 6)^2 + (y + 4)^2 = -16 + 36 + 16


(x - 6)^2 + (y + 4)^2 = 36.





A) Center is located at (6, -4).


B) The radius is 36^.5 = 6 units.





D) I'd help you with the length of AB, but you didn't even mention A or B in your answer.