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How do you write the equation of the circle in standard form, identify the center and radius of #x^2+y^2-2x+6y+9=0#?

Answer 1

Sophie Adams

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

#"the equation of a circle in standard form is "#

#color(red)(bar(ul(|color(white)(2/2)color(black)((x-a)^2+(y-b)^2=r^2)color(white)(2/2)|)))#

#"where "(a,b)" are the coordinates of the centre and r"# #"is the radius"#

#"to obtain this form use "color(blue)"completing the square"# #"on both x and y terms"#

#x^2-2x+y^2+6y+9=0#

#rArr(x^2+2(-1)xcolor(red)(+1)color(red)(-1))+(y^2+2(3)ycolor(blue)(+9)color(blue)(-9))+9=0#

#rArr(x-1)^2+(y+3)^2=0-9color(blue)(+9)color(red)(+1)#

#rArr(x-1)^2+(y+3)^2=1larrcolor(red)"in standard form"#

#rArr"centre "=(1,-3)" and radius "=1#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.