Let $z:\mathbb{R}^2\rightarrow \mathbb{R}$ be differentiable at $(x_{0},y_{0})$. The plane in $\mathbb{R}^3$ defined by the equation $T_{p}=z(x_{0},y_{0})+\left[\frac{\partial z}{\partial x}(x_{0},y_{0})\right](x-x_{0})+\left[\frac{\partial z}{\partial y}(x_{0},y_{0})\right](y-y_{0})$ is called the tangent plane of the graph of $z$ at the point $(x_{0}, y_{0})$.
This is the definition, now we have to find the tangent plane for $z=x^2+y^2$ at the $(1,2)$.
$z(1,2)=5;$ $\left[\frac{\partial z}{\partial x}(1,2)\right](x-1)=2(x-1)=2x-2 ;$ $\left[\frac{\partial z}{\partial y}(1,2)\right](y-2)=4(y-2)=4y-8 .$
So, the equation for tangent plan at point $(1,2)$ is :
$T_{p}=5+2x-2+4y-8=2x+4y-5. $