Prove that the square of the distance between the two points (x1,y1) and (x2,y2) of the circle x^2+y^2=a^2 is 2(a^2-x1x2-y1y2)?

2 Answers
Feb 15, 2018

#d^2=2(a^2-x_1x_2-y_1y_2)# (Proved)

Explanation:

Equation of circle is #x^2+y^2=a^2; (x_1,y_1) and (x_2,y_2)#

are two points on the circle #:. x_1^2+y_1^2=a^2# and

#: x_2^2+y_2^2=a^2# . Let the distance between two points

be #d# , then #d^2=(x_2-x_1)^2 +(y_2-y_1)^2#

#:. d^2=x_2^2-2x_1x_2+x_1^2+y_2^2-2y_1y_2+y_1^2# or

#d^2=x_1^2+y_1^2 +x_2^2+y_2^2-2x_1x_2-2y_1y_2#

Putting #:. x_1^2+y_1^2=a^2 and x_2^2+y_2^2=a^2# we get

#d^2=a^2+a^2-2x_1x_2-2y_1y_2# or

#d^2=2a^2-2x_1x_2-2y_1y_2# or

#d^2=2(a^2-x_1x_2-y_1y_2)# (Proved)

Feb 15, 2018

see below

Explanation:

We know,
Distance between any two points is
#d^2#=#(x_2-x_1)^2#+#(y_2-y_1)^2#
{By applying #color(red)("Distance formula")#}

but #x_1,x_2# and #y_2,y_1# lie on the circle whose equation is #x^2 +y^2=a^2#

Therefore,#(x_1)^2 +(y_1)^2=a^2#------#color(red)2#
and, #(x_2)^2 +(y_2)^2=a^2#--------------#color(red)3#
Substituting,we get,
#d^2#=#(x_2)^2+(x_1)^2 -2x_2x_1#+#(y_2)^2+(y_1)^2 -2y_1y_2#
or,
#a^2 +a^2-2(x_1x_2+y_1y_2)#
or,#2(a^2-x_1x_2-y_1y_2)#=#d^2#

# color(blue)(Hence Proved)#