How do you find the center and radius of the circle #x^2 + y^2 - 4x - 14y + 29 = 0#?

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Answer 1 · 2018-04-15T08:19:14

(2,7) centre and radius #2sqrt6#

The general equation of a circle is #(x-a)^2+(y-b)^2=r^2#

Where the centre is #(a,b) # and the radius is #r#

So we need to use completing the square and then rearrange the equation

#x^2-4x+y^2-14y=-29#

#[(x-2)^2-4]+[(y-7)^2-49]=-29#

#(x-2)^2+(y-7)^2-53=-29#

#(x-2)^2+(y-7)^2=-29+53#

#(x-2)^2+(y-7)^2=24#

So the centre is (2,7) and radius is #sqrt24# or #2sqrt6#