Find the equation of a circle passing through (3,7) and tangent to the line x-3y+8=0

Answer 1

The the centre be C(a,b)
and the point of contact with the line P(x,y)
let A(3,7) be the given point on the circle

AP must be a diameter, and at right angles to the given line
slope of given line = 1/3
so slope of diameter = -3
but (3,7) lies on it, so the equation is
y = -3x + b
7 = -3(3) + b
b = 16
diameter has equation y = -3x + 16

solving the two equations would give us P
x - 3(-3x+16) + 8 = 0
x + 9x - 48 + 8 = 0
10x = 40
x = 4
then y = -3(4) + 16 = 4
P is the point (4,4)
so the centre must be the midpoint and
C = (7/2, 11/2)

circle equation:
(x-7/2)^2 + (y-11/2)^2 = r^2
r = 1/2 the diameter
diameter = √( (-1)^2 + 3^2 ) = √10
r = √10/2
r^2 = 10/4 = 5/2

circle equation:
(x-7/2)^2 + (y-11/2)^2 = 5/2