Just a few points to add!
From your circle equation you should be able to recognise a few things. Firstly we know the center of the circle is the origin as it has not been translated by anything if it had it would look like (x - 2)^2 + (y - 5)^2 = 13 or similar.
Secondly any tangent of a circle will
ALWAYS make a right angle with the radius at the point where it touches the circle. With that in mind the equation of the normal is simply the equation of the radius at that point.
So we have the coords for the center of the circle (0, 0) and we have the the coords for the point on the circumference (-2, 3) so with these we can find the gradient of the radius thus m = (3 - 0)/(-2 - 0) ie change in y divided by the change in x and we get m = - 3/2. To find the equation of the normal we need the gradient of the normal which we simply use the rule
m1 x m2 = -1. That is if two lines are at right angles then the product of there gradients = -1. so -3/2 x m2 = -1 m2 = -1/-3/2 m2 = 2/3 So the gradient of the normal is 2/3. so the eqation of the normal is worked out thus
2x + 4 = y – 3
3 3 NOTE - This should read 4/3 but i cant get the 3 under the 4!
2x + 4 = 3y – 9
2x + 13 = 3y
I would leave it as 3y – 2x = 13
Hope that adds to the above post!