How to find the standard form of the equation of the specified circle given that it Passes through points A(-1,3) y B (7,-1) and it center is on line 2x+y-11=0?

1 Answer
Aug 8, 2016

Answer:

#(x-4)^2+(y-3)^2=5^2#

Explanation:

Let the coordinate of the center of the circle be #(a,b)" and its radius be " r #

So the equation of the circle is

#color(red)(t(x-a)^2+(y-b)^2=r^2)#
By the given condition its center lies on the straight line #2x+y-11=0#

So

#2a+b=11......(1)#

Again #A(-1,3) and B(7,-1)# are on the circle. So

#(a+1)^2+(b-3)^2= (a-7)^2+(b+1)^2#

#=>(a+1)^2+(b-3)^2-(a-7)^2-(b+1)^2=0#

#=>8(2a-6)-4(2b-2)=0#

#=>(2a-6)-(b-1)=0#

#=>2a-6-b+1=0#

#=>2a-b=5.....(2)#

Subtracting (2) from (1) we get

#2b=6=>b=3#

putting the value of b in (2)

#2a-3=5=>a=4#

Radius of the circle

#r=sqrt((a+1)^2+(b-3)^2)#

#=sqrt((4+1)^2+(3-3)^2)=5#

So equation of the circle

#(x-4)^2+(y-3)^2=5^2#