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.