Geometry

Is this what I am doing so far right?
(x-1)^2 + (y-3)^2 = 2((x+1)^2 + (y-2)^2)

Not quite.

The distance $d_1$ from $P_1$ to $(1,3)$ is given by $(x-1)^2+(y-3)^2=d_1^2$

Then distance $d_2$ from $P_1$ to $(-1,2)$ is given by $(x+1)^2+(y-2)^2=d_2^2$

We know that $d_1=2d_2$

Carry on from there.

Not quite.

The distance $d_1$ from $P_1$ to $(1,3)$ is given by $(x-1)^2+(y-3)^2=d_1^2$

Then distance $d_2$ from $P_1$ to $(-1,2)$ is given by $(x+1)^2+(y-2)^2=d_2^2$

We know that $d_1=2d_2$

Carry on from there.

With every example will I have to square root?

Not sure what you mean. I would avoid square roots.

So I expanded d1^2= x^2-2x+y^2-6y+10
and expanded d2^2=x^2+2x+y^2-4y+5
What do I do next?

Merge the two somehow, using the fact how the two distances are related to each other.

You are aiming to end up with some sort of equation in terms of $x$ and $y$ (which is what is meant by Cartesian coordinates in your OP) and no $d$ involved.

since d1=2d2 right can we do d1^2=4d2^2 and then replace the d1^2 with the 4d2^2?

Yep.

What about this Q
The distance from P4 to the point (-5,3) is twice the distance from P4 to the line with equation 4x+3y+5=0

@RDKGames
Well Ik d1=2d2
which means d1^2=4d2^2
but how do i get d2?

A point on the line has coordinates $(x,-\frac{1}{3}(4x+5))$

Is that d2?

No, that's a point on the line.

$d_2$ would be the distance between $P_1$ and this point.

Though of course, since we are saying $P_1$ has coordinates $(x,y)$, it would be helpful to instead say that the point on the line is in fact $(p,-\frac{1}{3}(4p+5))$ for some $p \in \mathbb{R}$

Ohh okayy yup! May I have a hint how to work out d2? Ive worked out d1 but dont know how to work out d2
Btw thank you so much for helping me

Same method to work out $d_2$, just that you'd have your equation with a $p$ term.

Well the distance $d_1$ from $P_1$ to $(-5,3)$ is given by $(x+5)^2+(y-3)^2=d_1^2$

I dont get how I'd use the same method? Like how am I supposed to get the values?