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

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

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

A= 1 , B= 1 , D=0 , E = -2 , F = 0

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