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

Statement \(1\) : The only circle having radius \(\sqrt {10} \) and a diameter along line \(2x + y = 5\) is \(x^2 + y^2 - 6x +2y = 0\).
Statement \(2\) : \(2x + y = 5\) is a normal to the circle \(x^2 + y^2 -6x+2y = 0\).

  1. A Statement \(1\) is false; Statement \(2\) is true
  2. B Statement \(1\) is true; Statement \(2\) is true, Statement \(2\) is a correct explanation for Statement \(1\).
  3. C Statement \(1\) is true; Statement \(2\) is false
  4. D Statement \(1\) is true; Statement \(2\) is true; Statement \(2\) is not a correct explanation for Statement \(1\).
Verified Solution

Answer & Solution

Correct Answer

(A) Statement \(1\) is false; Statement \(2\) is true

Step-by-step Solution

Detailed explanation

Circle: \({x^2} + {y^2} - 6x - 2y = 0\,\,\,\,\,\,\,\,\,......\left( i \right)\) Line: \(2x + y = 5\,\,\,\,\,\,\,\,\,\,\,.......\left( {ii} \right)\) Center \( = \left( {3, - 1} \right)\) Now, \(2 \times 3 - 1 = 5\), hence center lies on the given line. Therefore line passes…
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