JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let the circle \(S: 36 x^{2}+36 y^{2}-108 x+120 y+C=0\) be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, \(x-2 y=4\) and \(2 x-y=5\) lies inside the circle \(S\), then :
- A \(100\,<\,\mathrm{C}\,<\,156\)
- B \(\frac{25}{9}\,<\,C\,<\,\frac{13}{3}\)
- C \(81\,<\,\mathrm{C}\,<\,156\)
- D \(100\,<\,\mathrm{C}\,<\,165\)
Answer & Solution
Correct Answer
(A) \(100\,<\,\mathrm{C}\,<\,156\)
Step-by-step Solution
Detailed explanation
\(S: 36 x^{2}+36 y^{2}-108 x+120 y+C=0\) \(\Rightarrow x^{2}+y^{2}-3 x+\frac{10}{3} y+\frac{C}{36}=0\) \(\text { Centre } \equiv(-g,-f) \equiv\left(\frac{3}{2}, \frac{-10}{6}\right)\) \(\text { radius }=r=\sqrt{\frac{9}{4}+\frac{100}{36}-\frac{C}{36}}\) Now,…
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