MHT CET · Maths · Circle
If the line \(3 x-4 y-7=0\) and \(2 x-3 y-5=0\) pass through diameters of a circle of area \(49 \pi\) square units, then the equation of the circle is
- A \(x^2+y^2-2 x+2 y-47=0\)
- B \(x^2+y^2+2 x-2 y-51=0\)
- C \(x^2+y^2-2 x+2 y+51=0\)
- D \(x^2+y^2+2 x+2 y+47=0\)
Answer & Solution
Correct Answer
(A) \(x^2+y^2-2 x+2 y-47=0\)
Step-by-step Solution
Detailed explanation
Centre of the circle is the point of intersection of the diameters
\(3 x-4 y-7=0\) and \(2 x-3 y-5=0\)
which is \((1,-1)\) and \(r=7\)
\(\begin{aligned} & (x-1)^2+(y+1)^2=7^2 \\ & \Rightarrow x^2+y^2-2 x+2 y-47=0\end{aligned}\)
\(3 x-4 y-7=0\) and \(2 x-3 y-5=0\)
which is \((1,-1)\) and \(r=7\)
\(\begin{aligned} & (x-1)^2+(y+1)^2=7^2 \\ & \Rightarrow x^2+y^2-2 x+2 y-47=0\end{aligned}\)
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