How do you find general form of circle tangent to the y-axis and has a center of (-6, 7)?

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Answer 1 · 2015-12-31T00:09:43

Use a little reasoning to find:

#(x+6)^2+(y-7)^2 = 6^2#

or if really fussy:

#(x-(-6))^2+(y-7)^2 = 6^2#

The general equation of a circle with centre #(h, k)# and radius #r# is:

#(x-h)^2+(y-k)^2 = r^2#

In our case #(h, k) = (-6, 7)# and the radius #r# must be #6# for the circle to touch the #y# axis.

graph{((x-(-6))^2+(y-7)^2-6^2)((x-(-6))^2+(y-7)^2-0.012)=0 [-24.75, 15.25, -3.52, 16.48]}