# 2002 STEP 1 Q1

Why is it I cannot multiply the first equation by 25 and the second by 18 and then equate the RHS of each equation to find the equation of the points of intersection?

$25(x+2)^2 + 50y^2 = 162(x-1)^2 + 288y^2$

$0 = 137x^2 - 424x + 238y^2 + 62$

The above is not in the form stated in the question. Why does eliminating y work but equating constants does not work in solving this problem?

NB: The problem is also Q17 in Advanced problems in core mathematics

### Interesting question

What do you mean by 'the equation of the points of intersection'? How many points of intersection are there?

Why do you think that this method should work?

### Realised my mistake

Apologies for the bad wording – I meant to say what the question asked – ‘equation of any circle passing through the points of intersection’
I think the flaw in the my idea is that equating constants can be used to find the points of intersection of two curves (inconvenient method in this case) like the example below:
$x^2 + y^2 = 4$
$x^2 + (y-3)^2 = 4$
However, my misunderstanding was that this will lead me to the equation of a circle through these points of intersection – as you pointed out, this does not work

### Equating constants

I'm not sure about this 'equating constants' method. Consider a simple example which is easy to sketch like
y - x = 1
x^2 + y^2 = 1
What are the intersection points of these curves? What equation do you get by equating constants, and what are the solutions of this equation - does it help to find the intersection points?

### Another view

Perhaps it's easier to think about what this means rather than a specific case. Suppose we have two equations
$f(x, y) = c$ and
$g(x, y) = c$
(where $f$ and $g$ are functions and $c$ is constant) and we want to find the points of intersection (i.e. the pairs of values $(x, y)$ with both $f(x, y) = c$ and $g(x, y) = c$).

If we 'equate constants' and set $f(x, y) = g(x, y)$, then this gives us all points $(x, y)$ where the expressions $f(x, y)$ and $g(x, y)$ are equal, but their common value isn't necessarily $c$! We could have $f(x, y) = d$ and $g(x, y) = d$ where $d$ is some other constant. In other words, the condition $f(x, y) = g(x, y)$ is necessary for $(x, y)$ to be a point of intersection, but not sufficient. The set of points satisfying $f(x, y) = g(x, y)$ can be strictly larger than the set of points satisfying $f(x, y) = c = g(x, y)$. This means that in general, it isn't so useful for finding the intersection points.

### Thanks - I understand your

Thanks - I understand your explanation as to why the 'equating constants' method may not be so useful in finding intersection points

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