Question

Question #9691f

Answer 1

`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

Answer 2

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`

Answer 3

`"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)`