So, I've attached ye question and have completed the square to solve it. However, what do I do after doing so , what can I do to find the answer?<br>
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Also for part b, how do I find the point of intersection?
You have the right line of thought when completing the square, however you have added to the equation without subtracting the equivalent amount to keep the equation the same. I would write:
$\displaystyle x^2-6x+y^2-4y-12=0$
$\displaystyle x^2-6x+9+y^2-4y+4=12+9+4$
$\displaystyle (x-3)^2+(y-2)^2=5^2$
Now, it's easy to see the circle is centered at (3,2) and has radius 5.
For part b) use $\displaystyle y=1$ and solve for $\displaystyle x$. You will have a quadratic in $\displaystyle x$ so there will be two roots, giving you 2 points, both with $\displaystyle y$-coordinate of 1.