Two Tangents intersect at Station $12+00.$ The back tangent has a bearing of N$45$oE and is $1,000$ feet long $($T $=$ Tangent Distance $=1,000$ feet$).$ The forward tangent has a bearing of S$45$oE and is $1,000$ feet long. A curve with a radius of $1,000$ feet is used to connect the two tangents. The Point of Intersection $($PI$)$ of the tangents is at coordinates $1,000,1,000.$ Using the length of the curve used to connect the Tangents, what is the station of the Point of Curvature $($PC$).$

PC Station $=$ PI Station $$ T $($Tangent distance$)$

One station $(1+00)=100$ feet$),1,000$ feet $=10+00$ or $10$ stations