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Problem 2

# Show that the square of the distance berween the two points $\left(x_{1}, y_{1}\right)$$and$$\left(x_{2}, y_{2}\right)$ on the circle $$x^{2}+y^{2}=a^{2}$$ is equal to $$2\left(a^{2}-x_{1} x_{2}-y_{1} y_{2}\right)$$

Expert verified
The square of the distance between the two points on the circle is indeed equal to $$2\left(a^{2}-x_{1} x_{2}-y_{1} y_{2}\right)$$, as we have proved by manipulation of the equation of a circle and the formula for calculating the square of the distance between two points in a 2D space.
See the step by step solution

## Step 1: Understand the Problem

The goal is to prove that the square of the distance between two points $$\left(x_{1}, y_{1}\right)$$ and $$\left(x_{2}, y_{2}\right)$$ on the circle, given by the equation $$x^{2}+y^{2}=a^{2}$$, is equal to $$2\left(a^{2}-x_{1} x_{2}-y_{1} y_{2}\right)$$.

## Step 2: Calculate square of distance between two points

The square of the distance $$d$$ between two points $$\left(x_{1}, y_{1}\right)$$ and $$\left(x_{2}, y_{2}\right)$$ in a 2D space is given by $$d^{2} = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$$. This can be rewritten as $$d^{2} = x_{1}^{2}+y_{1}^{2} + x_{2}^{2}+y_{2}^{2} - 2(x_{1} x_{2} + y_{1} y_{2})$$.

## Step 3: Substitute values

We know both $$\left(x_{1}, y_{1}\right)$$ and $$\left(x_{2}, y_{2}\right)$$ lie on the circle $$x^{2} + y^{2} = a^{2}$$, so we can substitute $$x_{1}^{2}+y_{1}^{2} = a^{2}$$ and $$x_{2}^{2} + y_{2}^{2} = a^{2}$$ into the equation. Doing this, we get: $$d^{2} = 2a^{2} - 2(x_{1} x_{2} + y_{1} y_{2})$$. This equation can be rewritten as $$d^{2} = 2\left(a^{2}-x_{1} x_{2} - y_{1} y_{2}\right)$$, which is the expression that we were tasked to prove.

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