Find a solution to Laplace's equation that satisfies polar coordinates and show that any solution produces perpendicular lines. 7a. Find a solution of Laplace's equations $u_{xx}+u_{yy}=0$ of the form $u(x,y)=Ax^2+Bxy+Cy^2 (A^2+B^2+C^2 \neq 0)$ which satisfies the boundary condition $u(cos ( \theta),sin( \theta))=cos(2\theta)+sin(2\theta)$ for all points $(cos(\theta),sin(\theta))$ on the circle $x^2+y^2=1$
My Attempt: We need to find a solution to Laplace's equation in the form of $u(x,y)=Ax^2+Bxy+Cy^2 (A^2+B^2+C^2 \neq 0)$ that satisfies the polar coordinate conditions. 
Suppose our solution to Laplace's equation is $u(x,y) = \frac{1}{2}x^2+xy-\frac{1}{2}y^2$
By taking the $u_x$, $u_{xx}$ $u_y $ and $u_{yy}$, we have 
$u_x = x + y$
$u_{xx} = 1$
$u_y = x - y$
$u_{yy} = -1$
Therefore, $u_{xx}+u_{yy} = 0 \rightarrow 1-1 = 0$ and we have satisfied Laplace's equation. 
7b. Show that the graph of any solution $u(x,y)$ of Laplace's equation of the form in $(a)$ intersections the $xy$-plane  in a pair of perpendicular lines through $(0,0)$
I have some equations that I've found in that particular form which are the following:
$2x^2+4xy-2y^2$
$x^2+2xy-y^2$
$3x^2+6xy-3y^2$
$\frac{-1}{2}x^2+xy+\frac{1}{2}y^2$
So what I need to do is to graph these equations and show that they are intersecting the xy-plane with perpendicular lines. I know that a perpendicular line has a negative slope and on top of that it intersects whereas parallel lines are just two lines that don't cross in the graph. 
It must also follow the condition that I must be in a circle of radius 1 because 
$x^2+y^2 =1$ and we know that $x^2+y^2 =r^2$ from vector calculus, so $r^2 =1$ and the $r = 1$... well a +1 and a -1 to be exact because we need to add those signs after we take the square root. SO, the solution that should not only satisfy the equation for Laplace must also not leave the circle which is really small. So it could be that the numerical values of A B and C must be either 1 as in on the circle or $\frac{1}{2}$ meaning inside the circle.
So, maybe another solution might be $\frac{1}{4}x^2+\frac{1}{2}xy-\frac{1}{4}y^2$
$u_x = \frac{1}{2}x+\frac{1}{2}$
$u_{xx} = \frac{1}{2}$
$u_y = \frac{1}{2}x-\frac{1}{2}y$
$u_{yy} = -\frac{1}{2}$
$u_{xx}+u_{yy} = \frac{1}{2}-\frac{1}{2} = 0$
My question is how do I graph these solutions? Do I graph as normal by putting some values of x to obtain my y or some values of y to obtain my x and then plot the points? It's been a long time. The last time I graph was in vector calculus with the x,y,z. 
I want to use these equations
$u(x,y) = \frac{1}{2}x^2+xy-\frac{1}{2}y^2$ [original from 7a]
$\frac{1}{4}x^2+\frac{1}{2}xy-\frac{1}{4}y^2$
$x^2+2xy-y^2$

The graph of the function $u$ is the subset $\{(x,y,u(x,y)):x,y\in\Bbb R\}\subset\Bbb R^3$. The points on the $xy$-plane are precisely those with $z$-coordinate $0$. Hence the intersection with our graph with the $xy$-plane is the set of all points $(x,y)$ with $u(x,y)=0$. You have $u(x,y)=x^2+2xy-y^2=0$. Can you solve this? As a hint, complete the square on the first two terms, ignoring the third, and then proceed from there.
