Find the center and radius of circle $(x-3)^3+(y+4)^2=10$ ?

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Answer 1 · 2017-02-07T05:40:40

Coordinates of the centre of the circle are $(3,-4)$ and circle's radius is $sqrt10$ .

Perhaps you mean the equation to be $(x-3)^2+(y+4)^2=10$

As $(x-3)^2+(y+4)^2=10$ is equivalent to

$(x-3)^2+(y-(-4))^2=(sqrt10)^2$

or $sqrt((x-3)^2+(y-(-4))^2)=sqrt10$

The Left hand side shows that the distance of $(x,y)$ from point $(3,-4)$

and as the equation indicates that this is always $sqrt10$

Equation $(x-3)^2+(y+4)^2=10$ indicates that $(x,y)$ is always at a distance of $sqrt10$ from $(3,-4)$ .

Hence, coordinates of the centre of the circle are $(3,-4)$ and circle's radius is $sqrt10$ .

Find the center and radius of circle (x-3)^3+(y+4)^2=10(x-3)^3+(y+4)^2=10?