Question

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

Answer 1

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.

---