If the line x + y = 0 is tangent to the circle (x – λ)² + (y – β)² = 50, then (λ + β)² =

The correct answer is 100.

Here is the step-by-step mathematical explanation:

1. Identify the center and radius of the circle:
The given equation of the circle is:
${(x-\lambda )}^2+{(y-\beta )}^2=50$
Comparing this with the standard equation of a circle ${(x-h)}^2+{(y-k)}^2=R^2$, we find:
Center of the circle, $C=(\lambda ,\beta )$
Radius of the circle, $R=\sqrt{50}$

2. Condition for tangency:
A line is tangent to a circle if and only if the perpendicular distance from the center of the circle to the line is equal to the radius of the circle.
The given line is:
$x+y=0$

The formula for the perpendicular distance d from a point $(x_1,y_1)$ to the line $Ax+By+C=0$ is:
$d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$

Substituting the center $(\lambda ,\beta )$ and the line $1·x+1·y=0$ into the distance formula, we get:
$d=\frac{|\lambda +\beta |}{\sqrt{1^2+1^2}}=\frac{|\lambda +\beta |}{\sqrt{2}}$

Since the line is tangent to the circle, we set this distance equal to the radius $R=\sqrt{50}$:
$\left|\frac{\lambda +\beta }{\sqrt{2}}\right|=\sqrt{50}$

3. Solve for ${(\lambda +\beta )}^2$:
Multiply both sides of the equation by $\sqrt{2}$:
$|\lambda +\beta |=\sqrt{50}·\sqrt{2}$
$|\lambda +\beta |=\sqrt{100}$
$|\lambda +\beta |=10$

Squaring both sides to find ${(\lambda +\beta )}^2$:
${(\lambda +\beta )}^2={10}^2=100$