वृत x2 + y2 = 9 पर बिंदु (4, 0) से स्पर्शक की लम्बाई क्या होगी?

  1. √7 इकाई
  2. √6 इकाई
  3. √11 इकाई
  4. √17 इकाई

Answer (Detailed Solution Below)

Option 1 : √7 इकाई
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अवधारणा:

बाह्य बिंदु (x1, y1) से वृत x2 + y2 = a2 के स्पर्शक की लम्बाई \(\sqrt {x_1^2 + y_1^2 - {a^2}}\) होगी

गणना:

दिया गया है: वृत की समीकरण x2 + y2 = 9 और बिंदु (4, 0).

जैसा की हम जानते है बाह्य बिंदु (x1, y1) से वृत x2 + y2 = a2 के स्पर्शक की लम्बाई \(\sqrt {x_1^2 + y_1^2 - {a^2}}\) होगी

यहाँ , x1 = 4 , y1 = 0 और a2 = 9.

इसलिए स्पर्शक की लम्बाई  √7 इकाई है
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