a circle has the equation 3(x-5)^2_3y^2=12. Find the center and radius and graph the circle. find the intercepts, if any.
plz show work
4 answers
Shouldn't there be a + sign between the 2 and the 3? What you have written is not the equation of a circle.
Yes, my mistake.find the center, radius and graph of circle. then find intercepts
3(x-5)^2+3y^2=12
plz show work
3(x-5)^2+3y^2=12
plz show work
divide everybody by 3
(x-5)^2 + y^2 = 4
centre : (5,0), radius 2
clearly the circle can't reach the y-axis,
(let x= 0, ---> 75 + 3y^2 = 12
3y^2 = - .... , not possible)
for the x-intercepts , let y = 0
3(x-5)^2 = 12
(x-5)^2 = 4
x-5 = ± 2
x = 7 or x = 3
(x-5)^2 + y^2 = 4
centre : (5,0), radius 2
clearly the circle can't reach the y-axis,
(let x= 0, ---> 75 + 3y^2 = 12
3y^2 = - .... , not possible)
for the x-intercepts , let y = 0
3(x-5)^2 = 12
(x-5)^2 = 4
x-5 = ± 2
x = 7 or x = 3
Convert:
TT/3 rad to degrees
TT/3 rad to degrees
1 answer