Submitted by Heirio on Mon, 08/21/2017 - 16:10

So, the RHS of the answer is shown to be 5 - 4a. However, after going through the question and then checking what I did with the hints, I found the RHS to be 4 - 4a. Perhaps this is a mistake on my part, but I can't see where I've made any sort of mistake.

## Assignment 8

This was a given answer in a STEP question, so is unlikely to be wrong (whereas if it was an answer in our hints and solutions there is a possibility of a typo etc).

SPOILERS!!! OTHER READERS, PLEASE ONLY READ AFTER HAVING HAD AN ATTEMPT AT THIS QUESTION.

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The hints are a little short on detail, so here are a few steps for you to check. When I eliminate $y$ I get the quadratic equation $x^2-50x+98=0$. One of the two solutions of this gives complex $y$ values, so we only want $x=2$.

$x=2$ gives $y=\pm 1$, so the POI are at $(2,1)$ and $(2,-1)$. A sketch will now show you that the centre of the circle lies on the $x$-axis, so can be written as $(a,0)$.

Using $(x-a)^2+y^2=r^2$ and the point $(2,1)$ gives $r^2 = (2-a)^2+1^2=4-4a+a^2+1 = 5-4a+a^2$.

The equation of the circle is hence:

\[

(x-a)^2+y^2=5-4a+a^2

\]

Expanding the brackets and cancelling $a^2$ gives the required equation.

## Huh. What I did was that I

Huh. What I did was that I found the LHS and for the RHS (r^2), I found what I thought would be the radius of the circle.

The expression differed depending on whether or not a was greater than or less than 2. If it were greater than 2, then r = a - 2. If it were less than 2, then r = 2 - a. With r^2, these two expressions give the same result: a^2 - 4a + 4. I drew a diagram of this and it seems to make sense.

Where have I gone wrong?

## AHA!

AHA!

I've found where I went wrong!

The intersection points were at x co-ordinate two. When I wrote down that the midpoint for the chord between the intersections was (2,0), I used this as evidence that the centre was (a,0). However, I also wrote down that the circle cut the x axis at (2,0), which was incorrect.