AP EAMCET · Maths · Circle
If the segments of the straight lines \(x+y=6\) and \(x+2 y=4\) are two diameters of a circle passing through \((6,2)\), then the equation of that circle is
- A \(x^2+y^2-2 x-4 y-20=0\)
- B \(x^2+y^2+6 x-4 y-68=0\)
- C \(x^2+y^2-16 x+4 y+48=0\)
- D \(x^2+y^2+2 x-10 y-32=0\)
Answer & Solution
Correct Answer
(C) \(x^2+y^2-16 x+4 y+48=0\)
Step-by-step Solution
Detailed explanation
\(L_1 \equiv x+y=6\) and \(L_2 \equiv x+2 y=4\) Point of intersection of lines \(L_1\) and \(L_2\) is centre \(C\) of the circle. Solving \(L_1\) and \(L_2\) : \(\because \quad x+y=6 \Rightarrow x=6-y\) Then, \(x+2 y=4\)…
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