Question:
The equation to the tangents to the circle \(x^2 + y^2 = 4\), which are parallel to \(x + 2y + 3 = 0\), are:
वृत्त \(x^2 + y^2 = 4\) की स्पर्श रेखाओं के समीकरण, जो \(x + 2y + 3 = 0\) के समानांतर हैं, हैं:
Options:
- \(x – 2y = 2\)
- \(x + 2y = \pm 2\sqrt{3}\)
- \(x + 2y = \pm 2\sqrt{5}\)
- \(x – 2y = \pm 2\sqrt{5}\)
- \(x – 2y = 2\)
- \(x + 2y = \pm 2\sqrt{3}\)
- \(x + 2y = \pm 2\sqrt{5}\)
- \(x – 2y = \pm 2\sqrt{5}\)
Correct Answer: c)
Explanation:
A line parallel to \(x + 2y + 3 = 0\) can be written as:
\[x + 2y + k = 0\]
The perpendicular distance from the centre of the circle \((0, 0)\) to the tangent must equal the radius \(r = 2\):
\[\frac{|0 + 2(0) + k|}{\sqrt{1^2 + 2^2}} = 2 \implies \frac{|k|}{\sqrt{5}} = 2 \implies |k| = 2\sqrt{5} \implies k = \pm 2\sqrt{5}\]
Therefore, the equations of the tangents are:
\[x + 2y = \pm 2\sqrt{5}\]
Hence, the correct option is (c).
\(x + 2y + 3 = 0\) के समानांतर रेखा का समीकरण: \(x + 2y + k = 0\)।
केंद्र (0, 0) से रेखा की लंबवत दूरी त्रिज्या \(r = 2\) के बराबर होनी चाहिए:
\[\frac{|k|}{\sqrt{1^2 + 2^2}} = 2 \implies |k| = 2\sqrt{5} \implies k = \pm 2\sqrt{5}\]
अतः स्पर्श रेखाएँ: \(x + 2y = \pm 2\sqrt{5}\) हैं।
अतः सही विकल्प (c) है।