A curve C has parametric equations $$x = \frac{t^2 + 5}{t^2 + 1}$$ $$y = \frac{4}{t^2 + 1}$$ $$t \in \mathbb{R}$$ Show that all points on C satisfy $$(x - 3)^2 + y^2 = 4$$ - Edexcel - A-Level Maths Pure - Question 1 - 2021 - Paper 2

Start with the equation to show: $(x - 3)^2 + y^2 = 4$.

Substituting the parametric equations:

1. Substitute for $x$: $x - 3 = \frac{t^2 + 5}{t^2 + 1} - 3 = \frac{t^2 + 5 - 3(t^2 + 1)}{t^2 + 1} = \frac{-2t^2 + 2}{t^2 + 1} = \frac{2(1 - t^2)}{t^2 + 1}$.

2. Thus, $(x - 3)^2 = \left( \frac{2(1 - t^2)}{t^2 + 1} \right)^2 = \frac{4(1 - t^2)^2}{(t^2 + 1)^2}$.

Next, substitute for $y$: $y = \frac{4}{t^2 + 1}$

Then, $y^2 = \left( \frac{4}{t^2 + 1} \right)^2 = \frac{16}{(t^2 + 1)^2}$.

Now combine the results:
$\frac{4(1 - t^2)^2}{(t^2 + 1)^2} + \frac{16}{(t^2 + 1)^2} = \frac{4((1 - t^2)^2 + 4)}{(t^2 + 1)^2}$.

Simplifying produces: $\frac{4(1 - 2t^2 + t^4 + 4)}{(t^2 + 1)^2} = \frac{4(t^4 - 2t^2 + 5)}{(t^2 + 1)^2}$.

We want to show that this equals 4: $\frac{4(t^4 - 2t^2 + 5)}{(t^2 + 1)^2} = 4$, which simplifies back to the original equation.