1. ## question

i have 2 questions, the first one i could do, its the second i need help with. 1) determine the coordinates of the centre and the radius of the circle with equation $x^2 + y^2 - 6x + 8y = 0$. so i put it in its squared form (if thats what you call it. if not, any corrections would be appreciated) like so:
$(x - 3)^2 - 9 + (y + 4)^2 - 16 = 0$
$\implies (x - 3)^2 + (y + 4)^2 = 25$ so the centre must be $(3, -4)$ and the radius is 5

the second question which i need help with is:
find the coordinates of the points P and Q, where the line $x = 7$ intersects this circle. could someone show me in detail the way you would answer this

thanks for any help

2. Replace x in the equation of the circle and solve for y.

3. ok i tried that, here's what i did:
made x = 7 in the equation
$(7 - 3)^2 + (y + 4)^2 = 0 \implies 16 + y^2 + 8y + 16 = 0$
$\implies y^2 + 8y = -32 \implies y^2 + y = -\frac{32}{8}$ $\implies y^2 + y = -4 \implies y + y = \sqrt {-4} \implies y + y = -2 \implies y = -1$ which i think is right so the answer must be $(7, -1)$ but the book gives another answer as well as this which is $(7, -7)$ how would you come to this answer?

4. No, you're wrong.

$16+(y+4)^2=25\Rightarrow (y+4)^2=9\Rightarrow y+4=\pm 3$

If $y+4=3\Rightarrow y=-1$

If $y+4=-3\Rightarrow y=-7$

So the points of intersection are $(7,-1)$ and $(7,-7)$