Question

Consider a circle $(x - \alpha)^2 + (y - \beta)^2 = 50$, where $\alpha, \beta> 0$. If the circle touches the line $y + x = 0$ at the point $P$, whose distance from the origin is $4\sqrt{2}$, then $(\alpha + \beta)^2$ is equal to ....

Answer 1

The given circle is:
$(x - \alpha)^2 + (y - \beta)^2 = 50.$

The center of the circle is $C(\alpha, \beta)$, and the radius of the circle is:
$r = \sqrt{50} = 5\sqrt{2}.$

The circle touches the line $y + x = 0$ at point $P$. The perpendicular distance from the center $C(\alpha, \beta)$ to the line $y + x = 0$ is equal to the radius of the circle:
$\text{Distance from } C(\alpha, \beta) \text{ to the line } y + x = 0 = r.$

Using the formula for the perpendicular distance from a point to a line:
$\text{Distance} = \frac{|\alpha + \beta|}{\sqrt{1^2 + 1^2}} = \frac{|\alpha + \beta|}{\sqrt{2}}.$

Equating this to the radius:
$\frac{|\alpha + \beta|}{\sqrt{2}} = 5\sqrt{2}.$

Simplify to find $|\alpha + \beta|$:
$|\alpha + \beta| = 5\sqrt{2} \cdot \sqrt{2} = 10.$

Since $\alpha, \beta > 0$, we have:
$\alpha + \beta = 10.$

The square of $\alpha + \beta$ is:
$(\alpha + \beta)^2 = 10^2 = 100.$

The Correct answer is; 100