217k views
0 votes
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 =

2 Answers

4 votes

Answer:

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.

User Tushar Kotecha
by
7.1k points
2 votes

Answer:

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

User Aden
by
6.7k points