How do you write the standard equation for the circle passing through pts (8,-2), (6,2) and (3,-7)?

1 Answer
Dec 31, 2017

Answer:

The equation of circle is #(x-3)^2+(y+2)^2=25#

Explanation:

Let the points are #P(8,-2) , Q( 6,2) ,R(3,-7)# and Centre of

the circle is #(a,b)# and radius is #r#

The distance from center to the given 3 points are equal.

#(8-a)^2+(-2-b)^2=(6-a)^2+(2-b)^2=(3-a)^2+(-7-b)^2#

#(8-a)^2+(-2-b)^2=(6-a)^2+(2-b)^2#

#:. 64-16a+cancela^2+4+4b+cancelb^2= 36-12a+cancela^2+4-4b+cancelb^2# or

#-4a+8b=-28 or 2b-a= -7(1)# Similarly,

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

#:. 64-16a+cancela^2+4+4b+cancelb^2= 9-6a+cancela^2+49+14b+cancelb^2# or

#-10a-10b= =-10 or a+b=1(2)# Adding equation(1) and (2) we

get, #3b=-6 or b =-2:. a =1-b=1-(-2)=3 :.#

Centre is #(3,-2)# Radius is #r=|PC|=sqrt ((8-a)^2+(-2-b)^2)#

#:. r= sqrt ((8-3)^2+(-2+2)^2)=5#

The equation of circle is #(x-a)^2+(y-b)^2=r^2# or

#(x-3)^2+(y+2)^2=25# [Ans]