How do you write an equation of the circle that passes through (4,2) with center (0,1)?

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Answer 1 · 2016-10-02T06:44:49

#x^2+y^2-2y-16=0#

As the circle has center at #(0,1)#, if its radius is #r#, its equation is

#(x-0)^2+(y-1)^2=r^2#

As this passes through #(4,2)#

#(4-0)^2+(2-1)^2=r^2#

ot #r^2=4^2+1^2=16+1=17#

Hence equation of circle is #x^2+(y-1)^2=17# or

#x^2+y^2-2y+1=17# or #x^2+y^2-2y-16=0#
graph{x^2+y^2-2y-16=0 [-10.25, 9.75, -3.12, 6.88]}