How do you find the center and radius of the circle $x^{2} + y^{2} - 8 y = 9$ $x^2 + y^2 - 8y = 9$ ?

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Answer 1 · 2018-03-18T12:23:07

$centre (0, 4), radius = 5$ $"centre "(0,4), " radius "=5$

we complete the square

$x^{2} + y^{2} - 8 y = 9$ $x^2+y^2-8y=9$

$\to x^{2} + (y^{2} - 8 y + 4^{2}) - 4^{2} = 9$ $rarrx^2+(y^2-8y+4^2)-4^2=9$

$x^{2} + {(y - 4)}^{2} - 16 = 9$ $x^2+(y-4)^2-16=9$

$x^{2} + {(y - 4)}^{4} = 25$ $x^2+(y-4)^4=25$

cmp with the standard eqn

#"centre "(a,b)," radius "=r

${(x - a)}^{2} + {(y - b)} = r^{2}$ $(x-a)^2+(y-b)^"=r^2$

we have

$centre (0, 4), radius = 5$ $"centre "(0,4), " radius "=5$

How do you find the center and radius of the circle x2+y2−8y=9x^2 + y^2 - 8y = 9?