Question:
With reference to the circles \(x^2 + y^2 + 2ax + c = 0\) and \(x^2 + y^2 + 2by + c = 0\), the two circles touch each other if:
वृत्तों \(x^2 + y^2 + 2ax + c = 0\) और \(x^2 + y^2 + 2by + c = 0\) के संदर्भ में, दोनों वृत्त एक दूसरे को स्पर्श करते हैं यदि:
Options:
- \(c = \sqrt{a^2 + b^2}\)
- \(\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c}\)
- \(c = \frac{1}{a^2} + \frac{1}{b^2}\)
- \(c = \frac{1}{a^2 + b^2}\)
- \(c = \sqrt{a^2 + b^2}\)
- \(\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c}\)
- \(c = \frac{1}{a^2} + \frac{1}{b^2}\)
- \(c = \frac{1}{a^2 + b^2}\)
Correct Answer: b)
Explanation:
Radii are \(r_1 = \sqrt{a^2 – c}\) and \(r_2 = \sqrt{b^2 – c}\).
Distance between centres is \(d = \sqrt{a^2 + b^2}\).
Two circles touch each other when \(d = r_1 + r_2\):
\[\sqrt{a^2 + b^2} = \sqrt{a^2 – c} + \sqrt{b^2 – c}\]
Squaring both sides and simplifying yields the standard condition:
\[\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c}\]
Hence, the correct option is (b).
त्रिज्याएँ: \(r_1 = \sqrt{a^2 – c}\) तथा \(r_2 = \sqrt{b^2 – c}\)।
केंद्रों की दूरी: \(d = \sqrt{a^2 + b^2}\)।
स्पर्श करने की शर्त \(d = r_1 + r_2\) को वर्ग करके हल करने पर मानक संबंध प्राप्त होता है:
\[\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c}\]
अतः सही विकल्प (b) है।