What is the standard form of the equation of a circle given points: (7,-1), (11,-5), (3,-5)?

1 Answer
Jun 21, 2017

Answer:

Standard form of circle is #(x-7)^2+(y+5)^2=16#

Explanation:

Let the equation of circle be #x^2+y^2+2gx+2fy+c=0#, whose centre is #(-g,-f)# and radius is #sqrt(g^2+f^2-c)#. As it passes though #(7,-1)#, #(11,-5)# and #(3,-5)#, we have

#49+1+14g-2f+c=0# or #14g-2f+c+50=0# ......(1)

#121+25+22g-10f+c=0# or #22g-10f+c+146=0# ...(2)

#9+25+6g-10f+c=0# or #6g-10f+c+34=0# ......(3)

Subtracting (1) from (2) we get

#8g-8f+96=0# or #g-f=-12# ......(A)

and subtracting (3) from (2) we get

#16g+112=0# i.e. #g=-7#

putting this in (A), we have #f=-7+12=5#

and putting values of #g# and #f# in (3)

#6xx(-7)-10xx5+c+34=0# i.e. #-42-50+c+34=0# i.e. #c=58#

andequation of circle is #x^2+y^2-14x+10y+58=0#

and its centre is #(7,-5)# abd radius is #sqrt(49+25-58)=sqrt16=4#

and standard form of circle is #(x-7)^2+(y+5)^2=16#

graph{x^2+y^2-14x+10y+58=0 [-3.08, 16.92, -9.6, 0.4]}