As core 1 Equation of a circle- Sixth form

Right, well you know the general formula of a circle is:

(x-a)^2+(y-b)^2=r^2

So to begin with you have:

(x+3)^2+(y-1)^2=r^2

Then you need to find the distance between the two points, giving the radius of the circle. You just need phythagoras here:

Root: (x2-x1)^2+(y2-y1)^2

And as that is r^2, you don't need to root that at all.

So your final equation is:

(x+3)^2+(y-1)^2=127

I think, correct me if am wrong

Almost, 81 + 36 = 117 not 127

Also latex helps a lot in answering questions - I was a bit overwhelmed by it at first but it is really easy to use when you get used to it.

Okk i dont see where ive gone wrong, if someone could help?

the question is
1). P(6,7) is a point on a circle with its centre at (-3,1), find the equation of the normal to circle at P

So i found the gradient first and i got 2/3, i then put it into the forumla

y-7=2/3(x-6) i then x3 to get rid of the fraction and got 3(y-7)=2(x-6) and expanded it to get 3y-21=2x-12 and rearranged it to make y the subject and got the answer:

y=2/3x +9 but the asnwer book said it was y=2/3x+3

what did i do wrong?

is it just me or is the gradient of a normal -1/m

and the gradient here is 2/3 => gradient of normal is -3/2 ??? or am i just being silly