C1 Question

Really stuck with this 1 mark question ...

A circle with centre C (-5,6) touches the y axis as shown in the diagram

Find the equation of the circle in the form
$(x-a)^2 + (y-b)^2 = r^2$

i found that out ... $(x+5)^2 + (y-6)^2 = 5$,

BUT the next question says verify that the point P (-2,2) lies on the circle

i dont know how to work it out help pl0x

Sub x = -2 and y = 2 into the equation. It lies on the circle if the equation equals 5.

I expanded the brackets and then subed it LHS is -16 and RHS is -20

i found that out ... $(x+5)^2 + (y-6)^2 = 5$,

I expanded the brackets and then subed it LHS is -16 and RHS is -20

5 !!!!!!

(-2+5)^2=9
(2-6)^2=16
=25

(-2+5)^2=9
(2-6)^2=16
=25

Is that the correct method to use in the exams?