Question

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)?

Answer 1

`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)

Answer 2

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)`