7. The circle with center $p$ is externally tangent to $(x+1)^{2}+y^{2}=1$ and internally tangent to $x^{2}+y^{2}-2 x-8=0$. Then, the coordinates of the point $p$ that is farthest from the origin among all points satisfying the conditions are $\qquad$ _.
7. $(2,0)$

Point $p$ is such that the sum of its distances to the points $(-1,0)$ and $(1,0)$ is 4.
Therefore, the equation of the locus of point $p$ is $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1 \quad(x \neq-2)$. The point farthest from the origin is $(2,0)$.