ANSWER
P has coordinates
[tex](-13,28)[/tex]
EXPLANATION
We need to first of all get the centre of the circle.
So let us complete the squares as follows,
[tex] {x}^{2} + {y}^{2} + 6x - 14y = 483[/tex]
[tex] {x}^{2} + 6x + {y}^{2} - 14y = 483[/tex]
[tex] {x}^{2} + 6x +{3}^{2} + {y}^{2} - 14y + ( - 7) ^{2} = 483 + {3}^{2} + {( - 7)}^{2} [/tex]
[tex] {(x + 3)}^{2} +(y - 7) ^{2} = 483 + 9+ 49[/tex]
[tex] {(x + 3)}^{2} +(y - 7) ^{2} = 541[/tex]
We compare this to the general equation of the circle
[tex] {(x - a)}^{2} + {(y - b)}^{2} = {r}^{2} [/tex]
This implies that, the center of the circle is
[tex]C(-3,7).[/tex]
Let the coordinates of Q be
[tex](m,n).[/tex]
Since PQ is a diameter,the centre is the midpoint of PQ.
We now use the midpoint formula,
[tex](\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})[/tex]
This implies that,
[tex]C(-3,7) = ( \frac{m + 7}{2}, \frac{n - 14}{2} )[/tex]
This implies
[tex] - 3 = \frac{m + 7}{2} [/tex]
[tex]m + 7 = - 6[/tex]
[tex]m = - 13[/tex]
[tex]7 = \frac{n - 14}{2} [/tex]
[tex]14 = n - 14[/tex]
[tex]n = 28[/tex]
Therefore the coordinates of Q are
[tex](-13,28)[/tex]
