How do you find the center and radius of the circle given $x^2+(y-14)^2=34$ ?

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Answer 1 · 2016-11-26T06:30:51

Center is $(0,14)$ and radius is $sqrt34$ .

The equation $x^2+(y-14)^2=34$ can be written as

$(x-0)^2+(y-14)^2=(sqrt34)^2$

or $sqrt((x-0)^2+(y-14)^2)=sqrt34$

It is apparent that this represents locus of a point which moves so that its distance from point $(0,14)$ is always $sqrt34$ .

Hence $x^2+(y-14)^2=34$ is the equation of a circle whose center is $(0,14)$ and radius is $sqrt34$ .

How do you find the center and radius of the circle given x^2+(y-14)^2=34x^2+(y-14)^2=34?