How do you find the equation of a circle in standard form given C(5,6) and r=3?

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How do you write the general form given a circle that passes through the given points. P(-2, 0),...
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Answer 1 · 2017-01-29T06:38:50

#x^2+y^2-10x-12y+52=0.#

The Eqn. of a Circle having Centre at the point #(h,k)# and Radius

#r# is given by, #(x-h)^2+(y-k)^2=r^2#.

In our example, therefore, the reqd. eqn. is,

#(x-5)^2+(y-6)^2=3^2, i.e.,#

#x^2+y^2-10x-12y+52=0.#