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

Equation of two diameters of a circle are \(2 x-3 y=5\) and \(3 x-4 y=7\). The line joining the points \(\left(-\frac{22}{7},-4\right)\) and \(\left(-\frac{1}{7}, 3\right)\) intersects the circle at only one point \(P(\alpha, \beta)\). Then \(17 \beta-\alpha\) is equal to

  1. A \(2\)
  2. B \(4\)
  3. C \(6\)
  4. D \(7\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2\)

Step-by-step Solution

Detailed explanation

Centre of circle is \((1, -1)\) Equation of \(A B\) is \(7 x-3 y+10=0 \ldots\) \(....(i)\) Equation of \(\mathrm{CP}\) is \(3 x+7 y+4=0 \ldots\)\(......(ii)\) Solving \((i)\) and \((ii)\) \(\alpha=\frac{-41}{29}, \beta=\frac{1}{29} \quad \therefore 17 \beta-\alpha=2\)
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