TS EAMCET · Maths · Circle
If the lines \(2 x-3 y=5\) and \(3 x-4 y=7\) are two diameters of a circle of radius 7 , then the equation of the circle is
- A \(x^2+y^2+2 x-4 y-47=0\)
- B \(x^2+y^2=49\)
- C \(x^2+y^2-2 x+2 y-47=0\)
- D \(x^2+y^2=17\)
Answer & Solution
Correct Answer
(C) \(x^2+y^2-2 x+2 y-47=0\)
Step-by-step Solution
Detailed explanation
Since, the lines \(2 x-3 y=5\) and \(3 x-4 y=7\) are the diameters of a circle. Therefore, the point of intersection is the centre of the circle. On solving the given equations, we get \(x=1\) and \(y=-1\) ie, the centre of the circle. \(\therefore\) Required equation of circle…
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