How do you find the center, radius, and the x-and y-intercepts of the following circle #y^2+ x^2-12y +10x 57=0#?

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Answer 1 · 2018-04-28T06:33:31

Centre (-5,6) radius 2
With a radius of 2 it will not touch either axis

I assume you have missed a sign from in front of the 57. I'll guess it's +57, just change it if not.

#(x-a)^2+(y-b)^2=r^2# is the general formula for a circle where #(a,b)# are the cords of the centre and #r# is the radius

Rearrange and complete the square

#(x+5)^2-25+(y-6)^2-36=-57#

#(x+5)^2+(y-6)^2-61=-57#

#(x+5)^2+(y-6)^2=4#

Centre (-5,6) radius 2

With a radius of 2 it will not touch either axis