JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
If one of the diameters of the circle \(x^2+y^2-10 x+\) \(4 y+13=0\) is a chord of another circle \(C,\) whose center is the point of intersection of the lines \(2 x+\) \(3 y=12\) and \(3 x-2 y=5\), then the radius of the circle \(\mathrm{C}\) is
- A \(\sqrt{20}\)
- B \(4\)
- C \(6\)
- D \(3 \sqrt{2}\)
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
\( 2 x+3 y=12 \) \(3 x-2 y=5\) \(13 x=39\) \(x=3, y=2\) Center of given circle is \((5,-2)\) \(\text { Radius } \sqrt{25+4-13}=4\) \(\therefore \mathrm{CM}=\sqrt{4+16}=5 \sqrt{2}\) \(\therefore \mathrm{CP}=\sqrt{16+20}=6\)
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