hich is the equation of a circle with diameter AB with A(5,4) and B(-1,-4)?
possible answers;
(x-5)^2+(y-4)^2=10
(x+5)^2+(y+4)^2=100
(x-2)^2+y^2=25
(x+2)^2+y^2=5
Not sure
The answer is (x-2)2+y2=25 i just took it on my quiz and all the other answers are wrong!!! so i got it wrong at first!! but this is the correct one!
Okay, 9+16=25
so would my answer be (x-2)^2+(y^2)=25
I thiught you had to take the square rt of 25
This is what I have but not sure if right
(x-(-2)^2+(y-3)^2=5^2
(x+2)^2+(y-3)^2=25
Standard equation of a circle with a radius of r and centre (x0,y0) is:
(x-x0)²+(y-y0)²=r²
Centre of circle:
((5-1)/2, (4+(-4))/2)=(2,0)
Out of the four choices, there is only one that has a centre positioned at (2,0).
(hint: refer to the standard equation of the circle)
Radius, r:
Radius of circle = distance from centre to circumference
r = sqrt((5-2)^2+(4-0)^2)=5
This confirms the choice above.
(hint: refer to the standard equation of the circle)
I do not have the same answer as you.
If AB is the diameter, the centre is located at the mid-point between A and B.
The mid-point between A and B is
((Xa+Xb)/2, (Ya+Yb)/2)=(2,0)
I do not know how you got -2 and 3 for the centre of the circle. Can you explain?
I was looking at wrong problem for this one a I got (x+5)^2+(y+4)^2=100
Your answer is not correct.
Please explain how you got
(x+5)^2+(y+4)^2=100 as your answer.
Hint:
Centre of circle is at:
((5-1)/2, (4+(-4))/2)=(2,0)
(x+2)^2+y^2=5
1. Centre is at (2,0)
but
(x+2)^2+y^2=5
has a centre at (-2,0)
2. radius is 5, so the right hand side should be 5^2=25.