How do you find the center and radius for #5x^2 + 5y^2 +10x-30y+49=0#?

2 Answers
Nov 18, 2016

Answer:

The center of this circle is#" " (-1,3)" "# and radius #" "1/sqrt5#

Explanation:

#" "#
The general form of the equation of a circle is :
#" "#
#color(blue)((x-a)^2 + (y-b)^2 = r^2)#
#" "#
where #" "(a,b)" "#is the center of the circle and #r# is its radius.
#" "#
#" "#
In the given exercise we are asked to transform this standard form
#" "#
into the general form of the equation of a circle.
#" "#
#" "#
First , the coefficients of #" "x^2 " "and" "y^2" "# should be equal to #1#.
#" "#
Then, complete the square.
#" "#
#" "#
#5x^2+5y^2+10x-30y+49=0#
#" "#
#rArr(5x^2+5y^2+10x-30y+49)/5=0/5#
#" "#
#rArrx^2 + y^2 + 2x - 6y + 49/5 = 0#
#" "#
#rArr(x^2 + 2x + 1) + (y^2 -6y +9) +(49/5 - 1 -9)=0#
#" "#
#rArr(x^2 +2x +1)+ (y^2-6y+9) +(49/5-5/5-45/5)=0#
#" "#
#rArr(x^2 +2x +1)+ (y^2-6y+9) +(-1/5)=0#
#" "#
#rArr(x +1)^2+ (y-3)^2 = 1/5#
#" "#
The equation of the circle is: #" "(x +1)^2+ (y-3)^2 = (1/sqrt5)^2#
#" "#
#" "#
Hence, The center of this circle is#" " (-1,3)" "# and radius #" "1/sqrt5#

Nov 18, 2016

Answer:

centre #(-1,3)#

radius #r=sqrt(1/5)=sqrt5/5#

Explanation:

To find centre and radius we need to rearrange the equation into the form:

#(x-a)^2+(y-b)^2=r^2#

where; #(a,b)# are the co-ordinates of the centre and #r# is the radius.

#5x^2+5y^2+10x-30y+49=0#

Divide by# 5 # first

#x^2+y^2+2x-6y+49/5=0#

Now complete the square on both the #x# and #y# terms.

#(x^2+2x)+(y^2-6y)+49/5=0#

#(x^2+2xcolor(red)(+1^2))+(y^2-6ycolor(blue)(+3^2))color(red)(-1^2)color(blue)(-3^2)+49/5=0#

#(x+1)^2+(y-3)^2=1/5#

centre #(-1,3)#

radius #r=sqrt(1/5)=sqrt5/5#