How do you write the equation of the circle with center(1,-2) and passes through (6,-6)?

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Answer 1 · 2016-04-24T07:58:53

#(x-1)^2+(y+2)^2 = 41#

The equation of a circle with centre #(h, k)# and radius #r# may be written:

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

We are given #(h,k) = (1,-2)#, so the only unknown is #r^2#.

Since the circle passes through #(6, -6)#, the values #x=6#, #y=-6# will satisfy the equation and we find:

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

#= (6-1)^2+((-6)-(-2))^2#

#= 5^2+(-4)^2#

#= 25+16#

#= 41#

So the equation of our circle may be written:

#(x-1)^2+(y+2)^2 = 41#

graph{((x-1)^2+(y+2)^2-41)((x-1)^2+(y+2)^2-0.02)((x-6)^2+(y+6)^2-0.02) = 0 [-8.58, 11.42, -8.52, 1.48]}