How do you find the equation of the circle centered at (h,k) = (-4,-2) and passing through the point (-1,2)?

Question

2773 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-15T06:12:21

# x^2+y^2+8x+4x-5=0#

The reqd. circle passes thro. the pt. #P(-1,2)# & has Centre

#C(-4,-2)#.

Hence, its radius #r# is given by the dist. #CP#, where,

#r^2=(-1+4)^2+(2+2)^2=9+16=25#.

Therefore, the eqn. of circle is,

#(x+4)^2+(y+2)^2=25#

#:. x^2+y^2+8x+4x+16+4-25=0,#" i.e.,

# x^2+y^2+8x+4x-5=0#