TS EAMCET · Maths · Circle
If the equation of the circle which cuts each of the circles \(x^2+y^2=4, x^2+y^2-6 x-8 y+10=0\) and \(x^2+y^2+2 x-4 y-2=0\) at the extremities of a diameter of these circles is \(x^2+y^2+2 g x+2 f y+c=0\), then \(g+f+c=\)
- A 9
- B -9
- C 12
- D -12
Answer & Solution
Correct Answer
(B) -9
Step-by-step Solution
Detailed explanation
Since the circle \(x^2+y^2+2 g x+2 f y+c=0\) cuts each given circles at extremities of a diameter Then the common chords will pass through the centre of the respective circles. So, common chord of \(x^2+y^2=4\) and \(x^2+y^2+2 g x+\) \(2 f y+c=0\) is \(2 g x+2 f y+c+4=0\) passes…
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