Question:
If the straight line \(4x + 3y + \lambda = 0\) touches the circle \(2(x^2 + y^2) = 5\), then \(\lambda\) is:
यदि सरल रेखा \(4x + 3y + \lambda = 0\) वृत्त \(2(x^2 + y^2) = 5\) को स्पर्श करती है, तो \(\lambda\) का मान है:
Options:
- \(\frac{5\sqrt{5}}{2}\)
- \(5\sqrt{2}\)
- \(\frac{5\sqrt{5}}{\sqrt{2}}\)
- \(\frac{5\sqrt{10}}{2}\)
- \(\frac{5\sqrt{5}}{2}\)
- \(5\sqrt{2}\)
- \(\frac{5\sqrt{5}}{\sqrt{2}}\)
- \(\frac{5\sqrt{10}}{2}\)
Correct Answer: d)
Explanation:
The equation of the circle is \(x^2 + y^2 = \frac{5}{2}\).
Centre is \((0, 0)\) and radius is \(r = \sqrt{\frac{5}{2}}\).
The line touches the circle if the perpendicular distance from \((0, 0)\) equals \(r\):
\[\frac{|4(0) + 3(0) + \lambda|}{\sqrt{4^2 + 3^2}} = \sqrt{\frac{5}{2}} \implies \frac{|\lambda|}{5} = \frac{\sqrt{5}}{\sqrt{2}}\]
\[|\lambda| = \frac{5\sqrt{5}}{\sqrt{2}} = \frac{5\sqrt{10}}{2}\]
Hence, the correct option is (d).
वृत्त: \(x^2 + y^2 = 5/2\), केंद्र \((0, 0)\), त्रिज्या \(r = \sqrt{5/2}\)।
स्पर्श करने की शर्त: केंद्र से रेखा पर लंबवत दूरी = \(r\):
\[\frac{|\lambda|}{\sqrt{4^2 + 3^2}} = \sqrt{\frac{5}{2}} \implies |\lambda| = 5 \times \frac{\sqrt{10}}{2} = \frac{5\sqrt{10}}{2}\]
अतः सही विकल्प (d) है।