Question #9691f

3 Answers
Feb 5, 2018

#x^2+y^2-12x+6y+25=0#

#x^2+y^2-12x+6y+25+36 +9 - 36 -9=0#

#x^2-12x+36+y^2+6y +9 =-25+36 +9#

#(x-6)^2+(y+3)^2 =20#

#(x – h)^2 + (y – k)^2 = r^2#, where #(h, k)# is the center #r# is the radius.

so comparing with the standard equation, the center is #(6,-3)# and the radius is #sqrt20# units

Feb 5, 2018

The center of the circle is located at #" "# #(a,b)=(6,-3)#

The radius is #" "# #r=2\sqrt{5}#

Explanation:

#(x−a)^2+(y−b)^2=r^2# is the circle equation with a radius #r#, centered at #(a, b)#.

To convert the given equation of circle in the standard form, use the completing square techinque as in algebra here.

#x^2-12x+6y+y^2=-25#

#(x^2-12x)+(y^2+6y)=-25#

#(x^2-12x+36)+(y^2+6y)=-25+36#

#(x-6)^2+(y^2+6y)=-25+36#

#(x-6)^2+(y^2+6y+9)=-25+36+9#

#(x-6)^2+(y+3)^2=-25+36+9#

#(x-6)^2+(y+3)^2=20#

Re-write in standard form:

#(x-6)^2+(y-(-3))^2=(2\sqrt{5})^2#

Feb 5, 2018

#"radius "=2sqrt5," centre "=(6,-3)#

Explanation:

#"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 the x and y terms"#

#"adding the squared value to complete the squares to"#
#"both sides"#

#x^2+2(-6)xcolor(red)(+36)+y^2+2(3)ycolor(magenta)(+9)=-25color(red)(+36)color(magenta)(+9)#

#(x-6)^2+(y+3)^2=20larrcolor(blue)"in standard form"#

#"with "a=6,b=-3" and "r=sqrt20=2sqrt5#

#rArr"radius "=2sqrt5" and centre "=(6,-3)#