JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
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 ................
- A \(103\)
- B \(102\)
- C \(55\)
- D \(100\)
Answer & Solution
Correct Answer
(D) \(100\)
Step-by-step Solution
Detailed explanation
\( S:(x-\alpha)^2+(y-\beta)^2=50 \) \( C P=r \) \( \left|\frac{\alpha+\beta}{\sqrt{2}}\right|=5 \sqrt{2} \) \( \Rightarrow(\alpha+\beta)^2=100\)
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