Posted by **xavier** on Friday, January 25, 2013 at 9:13am.

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

- Analytic Geometry -
**Reiny**, Friday, January 25, 2013 at 10:02am
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**

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