Question:
The equation of the circle which touches the axes at a distance 5 from the origin is \(x^2 + y^2 – 2\alpha x – 2\alpha y + \alpha^2 = 0\). What is the value of \(\alpha\)?
वृत्त का समीकरण जो मूल बिंदु से 5 की दूरी पर अक्षों को स्पर्श करता है, \(x^2 + y^2 – 2\alpha x – 2\alpha y + \alpha^2 = 0\) है। \(\alpha\) का मान क्या है?
Options:
- 4
- 5
- 6
- 7
- 4
- 5
- 6
- 7
Correct Answer: b)
Explanation:
A circle touching both coordinate axes at a distance 5 from the origin has its centre at \((5, 5)\) and radius \(r = 5\).
Its equation is:
\[(x – 5)^2 + (y – 5)^2 = 5^2 \implies x^2 – 10x + 25 + y^2 – 10y + 25 = 25 \implies x^2 + y^2 – 10x – 10y + 25 = 0\]
Comparing this with \(x^2 + y^2 – 2\alpha x – 2\alpha y + \alpha^2 = 0\):
\[2\alpha = 10 \implies \alpha = 5\]
Hence, the correct option is (b).
दोनों अक्षों को 5 की दूरी पर स्पर्श करने वाले वृत्त का केंद्र (5, 5) तथा त्रिज्या 5 होती है:
\[(x – 5)^2 + (y – 5)^2 = 5^2 \implies x^2 + y^2 – 10x – 10y + 25 = 0\]
समीकरण \(x^2 + y^2 – 2\alpha x – 2\alpha y + \alpha^2 = 0\) से तुलना करने पर:
\[2\alpha = 10 \implies \alpha = 5\]
अतः सही विकल्प (b) है।