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

Question

3741 views around the world

You can reuse this answer
Creative Commons License

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]}