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

Question

2237 views around the world

You can reuse this answer
Creative Commons License

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#