Respuesta :
Answer:
The value of A is 1, D is 0, E is -2 and F is 0.
Step-by-step explanation:
The given equation is
[tex]Ax^2+Ay^2+Dx+Ey+F=0[/tex] ...(1)
The standard form of the circle is
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Where, (h,k) is the center of the circle and r is the radius.
[tex](x-0)^2+(y-1)^2=1[/tex]
[tex]x^2+y^2-2y+1-1=0[/tex]
[tex]x^2+y^2-2y=0[/tex]
It can be written as
[tex]x^2+y^2+0x-2y+0=0[/tex] ....(2)
On comparing (1) and (2) we get.
[tex]A=1[/tex]
[tex]D=0[/tex]
[tex]E=-2[/tex]
[tex]F=0[/tex]
Therefore the value of A is 1, D is 0, E is -2 and F is 0.
Answer:
A= 1 , B= 1 , D=0 , E = -2 , F = 0
Step-by-step explanation:
Given : General equation of circle as
[tex]Ax^2+By^2+Dx+Ey+F=0[/tex]
Also given Center of circle as (0, 1) and radius r = 1
We have to find the value of coefficient A , B , D, E and F.
Equation of circle having center at (h,k) and radius r is given by,
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Substitute values
h = 0 , k = 1 , r = 1
we get,
[tex]\Rightarrow (x-0)^2+(y-1)^2=(1)^2[/tex]
Using algebraic identity , [tex](a-b)^2=a^2-2ab+b^2[/tex] , we get
[tex]\Rightarrow x^2+y^2+1-2y=1[/tex]
Solving , we get,
[tex]\Rightarrow x^2+y^2-2y=0[/tex]
Comparing with given general form , we get,
A= 1 , B= 1 , D=0 , E = -2 , F = 0
