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

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Answer 1 · 2016-12-02T17:31:37

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

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.