# How do you write an equation for a circle whose center is at (0,0) and that is tangent to x+y=6?

Jun 6, 2016

${x}^{2} + {y}^{2} = 18$

#### Explanation:

The intersection of straight ${s}_{1} \to y + x = 6$ and it's normal ${s}_{2}$ passing by ${p}_{c} = \left\{{x}_{c} , {y}_{c}\right\} = \left\{0 , 0\right\}$ the circumference center, is also the tangency point between the circumference and the straight ${s}_{1}$.

The ${s}_{2}$ equation can be obtained knowing that ${m}_{1}$ is the ${s}_{1}$ declivity. Then

${s}_{2} \to y = {y}_{c} - \left(\frac{1}{m} _ 1\right) \left(x - {x}_{c}\right)$

but ${m}_{1} = {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{{s}_{1}} = - 1$

so

${s}_{2} \to y = 0 - \left(\frac{1}{- 1}\right) \left(x - 0\right) = x$

${s}_{1} \cap {s}_{2} \to \left\{\begin{matrix}y + x = 6 \\ y - x = 0\end{matrix}\right. \to \left\{{x}_{t} = 3 , {y}_{t} = 3\right\}$

where ${p}_{t} = \left\{{x}_{t} , {y}_{t}\right\}$ is the tangency point.

we know that $r = \left\lVert {p}_{c} - {p}_{t} \right\rVert = 3 \sqrt{2}$

${\left(x - {x}_{c}\right)}^{2} + {\left(y - {y}_{c}\right)}^{2} = {r}^{2}$
${x}^{2} + {y}^{2} = 18$