# 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)?

Feb 15, 2018

${d}^{2} = 2 \left({a}^{2} - {x}_{1} {x}_{2} - {y}_{1} {y}_{2}\right)$ (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 $\therefore {x}_{1}^{2} + {y}_{1}^{2} = {a}^{2}$ and

$: {x}_{2}^{2} + {y}_{2}^{2} = {a}^{2}$ . Let the distance between two points

be $d$ , then ${d}^{2} = {\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}$

$\therefore {d}^{2} = {x}_{2}^{2} - 2 {x}_{1} {x}_{2} + {x}_{1}^{2} + {y}_{2}^{2} - 2 {y}_{1} {y}_{2} + {y}_{1}^{2}$ or

${d}^{2} = {x}_{1}^{2} + {y}_{1}^{2} + {x}_{2}^{2} + {y}_{2}^{2} - 2 {x}_{1} {x}_{2} - 2 {y}_{1} {y}_{2}$

Putting $\therefore {x}_{1}^{2} + {y}_{1}^{2} = {a}^{2} \mathmr{and} {x}_{2}^{2} + {y}_{2}^{2} = {a}^{2}$ we get

${d}^{2} = {a}^{2} + {a}^{2} - 2 {x}_{1} {x}_{2} - 2 {y}_{1} {y}_{2}$ or

${d}^{2} = 2 {a}^{2} - 2 {x}_{1} {x}_{2} - 2 {y}_{1} {y}_{2}$ or

${d}^{2} = 2 \left({a}^{2} - {x}_{1} {x}_{2} - {y}_{1} {y}_{2}\right)$ (Proved)

Feb 15, 2018

see below

#### Explanation:

We know,
Distance between any two points is
${d}^{2}$=${\left({x}_{2} - {x}_{1}\right)}^{2}$+${\left({y}_{2} - {y}_{1}\right)}^{2}$
{By applying $\textcolor{red}{\text{Distance formula}}$}

but ${x}_{1} , {x}_{2}$ and ${y}_{2} , {y}_{1}$ lie on the circle whose equation is ${x}^{2} + {y}^{2} = {a}^{2}$

Therefore,${\left({x}_{1}\right)}^{2} + {\left({y}_{1}\right)}^{2} = {a}^{2}$------$\textcolor{red}{2}$
and, ${\left({x}_{2}\right)}^{2} + {\left({y}_{2}\right)}^{2} = {a}^{2}$--------------$\textcolor{red}{3}$
Substituting,we get,
${d}^{2}$=${\left({x}_{2}\right)}^{2} + {\left({x}_{1}\right)}^{2} - 2 {x}_{2} {x}_{1}$+${\left({y}_{2}\right)}^{2} + {\left({y}_{1}\right)}^{2} - 2 {y}_{1} {y}_{2}$
or,
${a}^{2} + {a}^{2} - 2 \left({x}_{1} {x}_{2} + {y}_{1} {y}_{2}\right)$
or,$2 \left({a}^{2} - {x}_{1} {x}_{2} - {y}_{1} {y}_{2}\right)$=${d}^{2}$

$\textcolor{b l u e}{H e n c e P r o v e d}$