How do you find the center and radius of a circle using a polynomial $x^{2} + y^{2} - 6 x + 10 y + 9 = 0$ $x^2 + y^2 - 6x + 10y + 9 = 0$ ?

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Answer 1 · 2018-05-24T12:15:04

Center is at $(3, - 5)$ $(3 , -5)$ and radius is $r = 5$ $r=5$ unit.

$x^{2} + y^{2} - 6 x + 10 y + 9 = 0$ $x^2+y^2-6 x +10 y +9 =0$ or

$(x^{2} - 6 x) + (y^{2} + 10 y) = - 9$ $(x^2 -6 x) +(y^2+10 y )= -9$ or

$(x^{2} - 6 x + 9) + (y^{2} + 10 y + 25) = 34 - 9$ $(x^2 -6 x +9) +(y^2+10 y +25 )=34 -9$ or

${(x - 3)}^{2} + {(y + 5)}^{2} = 5^{2}$ $(x-3)^2 +(y +5)^2=5^2$ . The center-radius form of the circle

equation is $(x – h)^2 + (y – k)^2 = r^2$ , with the center being at

the point $(h, k)$ and the radius being $r$ . Center is at

$(3 , -5)$ and radius is $r=5$ unit.

graph{x^2+y^2-6 x+10 y+9=0 [-20, 20, -10, 10]} [Ans]

How do you find the center and radius of a circle using a polynomial x^2 + y^2 - 6x + 10y + 9 = 0x2+y2−6x+10y+9=0x^2 + y^2 - 6x + 10y + 9 = 0?