How do you find the equation of a circle center is at (0, 0) and that is tangent to the line x + y = 8?

1 Answer
Dec 2, 2016

circle eqn." "x^2+y^2=32


Explanation:

the eqn. of the circle is of the form " "x^2+y^2=r^2

if " "x+y=8" " is a tangent, we can solve these simultaneously.

y=8-x" " substitute into the circle eqn.

x^2+(8-x)^2=r^2

x^2+64-16x+x^2=r^2

giving

2x^2-16x+(64-r^2)=0

a quadratic in ""x
if " "x+y=8" " is a tangent then the quadratic must have EQUAL roots

ie " "b^2-4ac=0

we have:" "256-4xx2xx(64-r^2)=0

256-512+8r^2=0

8r^2=256
r=sqrt32=4sqrt2

circle eqn." "x^2+y^2=32


a quick check

put" " r^2=32 into the quadratic

2x^2-16x+32=0

x^2-8x+16=0

(x-4)^2=0

x=4
from " "x+y=8," "y=4

tangent at " "(4, 4)

double check with " "x^2+y^2=4^2+4^2=32

consistent.