MHT CET · Maths · Circle
If the lines \(2 x-3 y=5\) and \(3 x-4 y=7\) are the diameters of a circle of area154sq \(\cdot\) units, then equation of the circle is \(\left(\right.\) Take \(\left.\pi=\frac{22}{7}\right)\)
- A \(x^2+y^2-2 x+2 y-47=0\)
- B \(x^2+y^2-2 x+2 y-49=0\)
- C \(x^2+y^2-2 x-2 y-47=0\)
- D \(x^2+y^2-2 x-2 y-49=0\)
Answer & Solution
Correct Answer
(A) \(x^2+y^2-2 x+2 y-47=0\)
Step-by-step Solution
Detailed explanation
Centre is point of intersection of \(2 x-3 y=5\) and \(3 x-4 y=7\)
\(i \cdot e \cdot(1,-1)\)
and radius \(r\) is such that \(\pi r^2=154 \Rightarrow r=7\)
Hence, the required equation is
\(\begin{aligned} & (x-1)^2+(y+1)^2=7^2 \\ \Rightarrow & x^2+y^2-2 x+2 y-47=0\end{aligned}\)
\(i \cdot e \cdot(1,-1)\)
and radius \(r\) is such that \(\pi r^2=154 \Rightarrow r=7\)
Hence, the required equation is
\(\begin{aligned} & (x-1)^2+(y+1)^2=7^2 \\ \Rightarrow & x^2+y^2-2 x+2 y-47=0\end{aligned}\)
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