Question 682628
I don't know how you are expected to solve the problem,
but here goes my drawing-heavy, wordy solution.
Here are
the circle {{{x^2+y^2=9}}} (in red),
the line {{{7x+y+3=0}}} <--> {{{y=-7x-3}}} (also in red), and
the perpendicular bisector of PQ, line {{{y=x+1}}} (in blue).
{{{drawing(300,300,-5,5,-5,5,
grid(1),
red(circle(0,0,3)),
red(line(-1.2,5.4,0.4,-5.8)),
blue(line(-5,-4,5,6))
)}}}
The equation of the circle shows that it has radius {{{3}}} and is centered at the origin.
The equation of the red line shows that it passes through (0,-3),
so we know that it crosses the circle at that point.
The red line also crosses the circle at (-42/50,144/50)=(-0.84,2.88).
That is not so easy, but substituting {{{y=-7x-3}}} into {{{x^2+y^2=9}}},
we get {{{x^2+(7x+3)^2=9}}} --> {{{50x^2+42x+9=9}}} --> {{{50x^2+42x=0}}} --> {{{(50x+42)x=0}}},
with solutions {{{x=0}}} and {{{x=-42/50}}}.
Then substituting {{{x=-42/50}}} into {{{y=-7x-3}}}, we get
{{{y=-7(-42/50)-3}}} --> {{{y=294/50-150/50}}} --> {{{y-144/50}}}
From the equation for the blue line, with {{{slope=1}}},
we know that it forms a {{{45^o}}} angle with the x-axis,
and so it forms that little isosceles right triangle with the x-axis and the y-axis.
That slope, and the symmetry makes the problem easy.
(I found it very easy after scratching my head for an hour).
The point where the blue and red lines intersect is (-1/2,1/2), and I'll call it point R.
The red line {{{y=-7x-3}}} contains R and Q,
so I can call it line RQ.
We know Q is on that line.
{{{drawing(300,300,-3.5,3.5,-3.5,3.5,
grid(0),
red(circle(0,0,3)),
red(line(-1.2,5.4,0.4,-5.8)),
blue(line(-5,-4,5,6)),
locate(-0.8,0.7,R)
)}}}
We still do not see line PQ, but we know that
line PQ is perpendicular to the blue line.
We also know that
line PQ passes through P (a point in the circle), and
through Q, a point in the red line {{{y=-7x-3}}}.
Point R, on the perpendicular bisector of PQ,
is at the same distance from P and Q, and
is the vertex of isosceles triangle PRQ.
That blue line perpendicular bisector contains the altitude of PRQ,
and bisects angle PRQ.
I could draw line PR (in green),
as the reflection of the red line RQ on the blue line,
because I know that PR and RQ make the same angle with the blue line.
I could even draw the bisector of the other angle formed by PR and RQ.
I'll draw that one as a black line.
{{{drawing(300,300,-3.5,3.5,-3.5,3.5,
grid(0),
red(circle(0,0,3)),
red(line(-1.2,5.4,0.4,-5.8)),
blue(line(-5,-4,5,6)),
green(line(-5,8/7,5,-2/7)),
line(-5,5,5,-5),
locate(-1,0.5,R)
)}}}
Because they are bisectors of the angles formed by the same red and green lines,
the blue and black lines are perpendicular.
The slope of the black line is {{{-1}}}, and the line goes through the origin.
It is all very symmetrical;
the slopes of the red and green lines are {{{-7}}},
and {{{-1/7}}} respectively.
I may be missing something, but I see two points that could be P:
(3,0), where the green line crosses the circle on the right side, and
(-144/50,42/50)=(-2.88,0.84), where the green line crosses the circle on the left side.
P=(3,0) would mean Q=(-1,4)
(Q is 3.5 units up and 0.5 left from R,
because P is 3.5 units right and 0.5 units down)
P=(-144/50,42/50) would mean Q=(-8/50,-94/50)
(both points are 119/50 in one direction and 17/50 in another from R)
{{{drawing(300,300,-5,5,-5,5,
grid(0),
red(circle(0,0,3)),
red(line(-1.2,5.4,0.4,-5.8)),
blue(line(-5,-4,5,6)),
green(line(-5,8/7,5,-2/7)),
locate(-1,1,R),
line(3,0,-1,4),line(-144/50,42/50,-8/50,-94/50),
locate(3.2,1,P[1]),locate(-1.6,4.5,Q[1]),
locate(-3.3,1.8,P[2]),locate(0,-1.3,Q[2])
)}}}