How do you write an equation for a circle given center $(-sqrt13,42)$ and passes through the origin?

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Answer 1 · 2016-10-29T08:52:53

$(x+sqrt(13))^2+(y-42)^2=1777$

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

where the centre is at $(a,b)$ with a radius of r.

So for this what we are missing is the radius but we have the centre and a point $(0,0)$

the distance between the center and the point is,

$(0,0)-(-sqrt(13),42)$

$(sqrt13,-42)$

Using Pythagoras's theorem this means that the radius is,

$r=sqrt(sqrt(13)^2+42^2)$

$r=sqrt(1777)$

so back to the general equation,

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

sub in,

$(x+sqrt(13))^2+(y-42)^2=(sqrt(1777))^2$

$(x+sqrt(13))^2+(y-42)^2=1777$

How do you write an equation for a circle given center (-sqrt13,42)(-sqrt13,42) and passes through the origin?