Let $P$ be a point on the circle $x^2+y^2=9,Q$ a point on the line $7x+y+3=0,$ and the perpendicular bisector of $PQ$ be the line $x-y+1=0.$Then, the coordinates of $P$ are

Let the coordinates of $P$ and $Q$ be$(h, k)$ and $\left(x_1,y_1\right)$ respectively. Then, 

       $h^2+k^2=9$                …(i)

and $7x_1+y_1+3=0$        …(ii)

Since PQ is perpendicular to the line $x-y+1=0.$

$\therefore \frac{y_1-k}{x_1-h}=-1\Rightarrow x_1+y_1=h+k$         …(iii)

The mid point of PQ lies on $x-y+1 = 0$

$\therefore \frac{x_1+h}{2}-\frac{y_1+k}{2}+1=0\Rightarrow x_1-y_1=-(h-k+2)$   …(iv)

From (iii) and (iv), we get $x_1=k-1,y_1=h+1$

Substituting x1 , y1 in (ii), we get

$h+7k=3\Rightarrow h=3-7k$

Putting the value of h in $h^2+k^2=9$, we get

$50k^2-42k=0\Rightarrow k=0,21/25.$

Hence, the coordinates of $P$ are $(3,0)$ or, $(-72/25,21/25)$