JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
If one of the diameter s of the circle, given by the equation, \({x^2} + {y^2} - 4x + 6y - 12 = 0\) is a chord of a circle \(S,\) whose centre is at \((-3, 2),\) then the radius of \(S\) is:
- A \(5\)
- B \(10\)
- C \(5\sqrt 2 \)
- D \(\;5\sqrt 3 \)
Answer & Solution
Correct Answer
(D) \(\;5\sqrt 3 \)
Step-by-step Solution
Detailed explanation
Given equation of a circle is \(x^{2}+y^{2}-4 x+6 y-12=0,\) whose centre is \((2,-3)\) and radius \(=\sqrt{2^{2}+(-3)^{2}+12}\) \(=\sqrt{4+9+12=5}\) Now, according to given information, we have the following figure. \(x^{2}+y^{2}-4 x+6 y-12=0\) clearly, \(AO\) perpendicular to…
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