TS EAMCET · Maths · Circle
Let \(S \equiv x^2+y^2-6 x-6 y+4=0\) and \(S^{\prime} \equiv x^2+y^2-2 x-4 y+3=0\) be two circles. The centre of a circle of radius \(\sqrt{14}\) and having same radical axes with either of \(S=0\) or \(S^{\prime}=0\) is
- A \((3,3)\)
- B \(\left(\frac{-19}{5}, \frac{-2}{5}\right)\)
- C \((1,2)\)
- D \(\left(\frac{2}{5}, \frac{3}{5}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{-19}{5}, \frac{-2}{5}\right)\)
Step-by-step Solution
Detailed explanation
Let the equation of required circle is \(x^2+y^2+2 g x+2 f y+c=0\) and the circle \(1, s=0\) and \(s^1=0\) are of a coaxial system of circles, so \(\frac{g+3}{2}=\frac{f+3}{1}=\frac{c-4}{-1}=k\) (let)…
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