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

Dec 2, 2016

circle eqn.$\text{ } {x}^{2} + {y}^{2} = 32$

#### Explanation:

the eqn. of the circle is of the form $\text{ } {x}^{2} + {y}^{2} = {r}^{2}$

if $\text{ "x+y=8" }$ is a tangent, we can solve these simultaneously.

$y = 8 - x \text{ }$ substitute into the circle eqn.

${x}^{2} + {\left(8 - x\right)}^{2} = {r}^{2}$

${x}^{2} + 64 - 16 x + {x}^{2} = {r}^{2}$

giving

$2 {x}^{2} - 16 x + \left(64 - {r}^{2}\right) = 0$

a quadratic in ""x
if $\text{ "x+y=8" }$ is a tangent then the quadratic must have EQUAL roots

ie $\text{ } {b}^{2} - 4 a c = 0$

we have:$\text{ } 256 - 4 \times 2 \times \left(64 - {r}^{2}\right) = 0$

$256 - 512 + 8 {r}^{2} = 0$

$8 {r}^{2} = 256$
$r = \sqrt{32} = 4 \sqrt{2}$

circle eqn.$\text{ } {x}^{2} + {y}^{2} = 32$

a quick check

put$\text{ } {r}^{2} = 32$ into the quadratic

$2 {x}^{2} - 16 x + 32 = 0$

${x}^{2} - 8 x + 16 = 0$

${\left(x - 4\right)}^{2} = 0$

$x = 4$
from $\text{ "x+y=8," } y = 4$

tangent at $\text{ } \left(4 , 4\right)$

double check with $\text{ } {x}^{2} + {y}^{2} = {4}^{2} + {4}^{2} = 32$

consistent.