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)