Find the standard form of the equation of the circle with center (-3 ,5 ) and tangent to the line y = 1.

Standard form of the equation of the circle is

(x-p) ∧2 + (y-q) ∧2 = r∧2 where the (p, q) = (-3,5) and r radius

Explicit form of the straight line is y=kx+n, where k is the line coefficient and n intersection with y axis. In our case k=0 and n=1

The equation of the touch condition between circle and line is

r∧2 (1+k∧2) = (kp-q+n) ∧2

When we replace all we know in this equation we get

r∧2 (1+0) = (0-5+1) ∧2 = > r∧2 = (-4) ∧2 = > r∧2=16

And equation of the circle is (x+3) ∧2 + (y-5) ∧2=16 or

x∧2+y∧2+6x-10y+18=0

Good luck!