Question

Find the general form, Ax2+Ay2+Dx+Ey+F=0, by identifying the coefficients A,D,E,&F center:(0,1); r=1 (x-0)2+(y-1)2=1 x2+y2-2y+1-1=0 x2+y2-2y=0 A= Blank D = Blank E= Blank F =

Answer 1

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

$Ax^2+Ay^2+Dx+Ey+F=0$             ...(1)

The standard form of the circle is

$(x-h)^2+(y-k)^2=r^2$

Where, (h,k) is the center of the circle and r is the radius.

$(x-0)^2+(y-1)^2=1$

$x^2+y^2-2y+1-1=0$

$x^2+y^2-2y=0$

It can be written as

$x^2+y^2+0x-2y+0=0$                          ....(2)

On comparing (1) and (2) we get.

$A=1$

$D=0$

$E=-2$

$F=0$

Therefore the value of A is 1, D is 0, E is -2 and F is 0.

Answer 2

Answer:

A= 1 , B= 1 , D=0 , E =  -2 , F = 0

Step-by-step explanation:

Given : General equation of circle as

$Ax^2+By^2+Dx+Ey+F=0$

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,

$(x-h)^2+(y-k)^2=r^2$

Substitute values

h = 0 , k = 1 , r = 1

we get,

$\Rightarrow (x-0)^2+(y-1)^2=(1)^2$

Using algebraic identity , $(a-b)^2=a^2-2ab+b^2$ , we get

$\Rightarrow x^2+y^2+1-2y=1$

Solving , we get,

$\Rightarrow x^2+y^2-2y=0$

Comparing with given general form , we get,

A= 1 , B= 1 , D=0 , E =  -2 , F = 0