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

Correct Answer: 100

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