How do you write the equation in standard form #x^2 + y^2 - 4x - 14y + 29 = 0#?

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Answer 1 · 2015-12-01T07:12:59

Complete the square for #x# and #y# to find:

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

or if you prefer:

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

which is in standard #(x-h)^2+(y-k)^2=r^2# form.

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

#=(x^2-4x+4)+(y^2-14y+49)+(29-4-49)#

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

Add #24# to both ends to get:

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

This is the equation of a circle with centre #(2, 7)# and radius #sqrt(24) = 2sqrt(6)#