Answered

Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

The equation x2 + y2 − 2x + 2y − 1 = 0 is the general form of the equation of a circle. What is the standard form of the equation?

Sagot :

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

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

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

(x - 1)^2 + (y + 1)^2 - 3 = 0

(x - 1)^2 + (y + 1)^2 = 3


Answer:

[tex](x-1)^2+ (y+1)^2=3[/tex]

Step-by-step explanation:

[tex]x^2 + y^2 - 2x + 2y - 1 = 0[/tex]

Standard form of the equation is [tex](x-h)^2 + (y-k)^2= r^2[/tex]

To get standard form we apply completing the square method

[tex]x^2-2x+ y^2+ 2y - 1 = 0[/tex]

Take coefficient of x  and y . Divide it by 2 and then square it

[tex]\frac{2}{2} =1[/tex] and 1^2=1

Add and subtract 1

[tex](x^2-2x)+(y^2+ 2y) - 1 = 0[/tex]

[tex](x^2-2x+1-1)+(y^2+ 2y+1-1) - 1 = 0[/tex]

[tex](x^2-2x+1)+(y^2+ 2y+1)-1-1- 1 = 0[/tex]

[tex](x^2-2x+1)+(y^2+ 2y+1)-3= 0[/tex]

Now write the parenthesis in square form

[tex](x-1)(x-1)+ (y+1)(y+1)-3= 0[/tex]

[tex](x-1)^2+ (y+1)^2-3= 0[/tex] , add 3 on both sides

[tex](x-1)^2+ (y+1)^2=3[/tex] is the standard form

We hope this was helpful. Please come back whenever you need more information or answers to your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.