Why am I not able to find the intersection of circles $(x-2)^2+(y-5)^2=5$ and $(x-1)^2+(y+3)^2=50$ algebraically?

Question

I'm answering a question on finding the intersection of two circles. I am confused why my algebraic method is not leading me to the correct answers.

The two circles have equations: $$(x-2)^2+(y-5)^2=5 \quad\text{and}\quad (x-1)^2+(y+3)^2=50$$

First, I expanded both and got:

$$\begin{align} x^2-4x+4+y^2-10y+25&=\phantom{1}5 \\ x^2-2x+1+y^2+\phantom{1}6y+\phantom{1}9&=50 \end{align}$$

I then multiplied the first equation by -1, and added it to the second equation to get:

$x=32-8y$

I know that everything up to here is correct, because as far as I'm aware, simultaneously solving them should tell me equation of the line that goes through their intersection points. I graphed the line and the two circles on Desmos and this confirmed the line was right. It also clearly shows one of the solutions as (0,4)

So now, surely just substitute the equation $x=32-8y$ into one of the circles to find the intersections?

I substituted it into the first one:

$(32-8y)^2-4(32-8y)+4+y^2-10y+25=5$

Which simplified to:

$65y^2-494y+920=0$

But solving this quadratic doesn't give a y-coordinate of 4 as a solution. It seems to give totally random solutions.

Where have I gone wrong?

Answer 1

Substituing $x=32-8y$ directly into

$$ (x-2)^2+(y-5)^2=5 $$ gives

$$ 65y^2 - 490y + 920 = 0$$

Answer 2

You simplified the quadratic wrong:\begin{eqnarray} (32 - 8y)^2 - 4(32-8y) + 4 + y^2 - 10y + 25 &=& 5\\ 1024 - 512 y + 64 y^2 - 128 + 32 y + 4 + y^2 - 10y + 25&=&5\\ (64+1) y^2 + (-512 + 32 - 10) y + (1024-128+4+25-5) &=& 0\\ 65 y^2 - 490 y + 920 &=&0 \end{eqnarray} The $y$ coefficient should be $-490$ not $-494$.

Answer 3

From the correct equation $$ 65y^2-490y+920=0 $$ You get 2 solutions: $y=4,46/13$. Subtituting in in one of the equation $x=32-8y$ gives $x=0,48/13$, respectively. Note that the equation of line that you got $x=32-8y$, describes correctly the line on which both points lie.