ExamBro
ExamBro
enEnglishhiहिन्दीguગુજરાતી
JEE Mains · Maths · STD 11 - 10.1 circle and system of circle

\(\lambda \) ની એવી શકય કિમતોનો ગણ મેળવો કે જેથી વર્તુળ \(x^2 + y^2 - 4x - 4y+ 6\, = 0\) અને \(x^2 + y^2 - 10x - 10y + \lambda \, = 0\) ને બરાબર બે સામાન્ય સ્પર્શકો હોય 

  1. A \((12, 32)\)
  2. B \((18, 42)\)
  3. C \((12, 24)\)
  4. D \((18, 48)\)
Verified Solution

Answer & Solution

Correct Answer

(B) \((18, 42)\)

Step-by-step Solution

Detailed explanation

The equation of the circles are \({x^2} + {y^2} - 10x - 10y + \lambda = 0\,\,\,\,\,\,\,......\left( 1 \right)\) and \({x^2} + {y^2} - 4x - 4y + 6 = 0\,\,\,\,\,\,\,......\left( 2 \right)\) \({C_1} = \,\) center of \(\left( 1 \right) = \left( {5,5} \right)\) \({C_2} = \,\) center…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app