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!!

ddd

step-by-step explanation: