Q52 / 120 · VITEEE 2012 · Mathematics
The lines $2 x-3 y-5=0$ and $3 x-4 y=7$ are diameters of a circle of area $154 \mathrm{sq}$ units, then the equation of the circle is
- A$x^{2}+y^{2}+2 x-2 y-62=0$
- B$x^{2}+y^{2}+2 x-2 y-47=0$
- Correct answerC$x^{2}+y^{2}-2 x+2 y-47=0$
- D$x^{2}+y^{2}-2 x+2 y-62=0$
Solution
The centre of the required circle lies at the intersection of $2 x-3 y-5=0$ and $3 x-4 y-$ $7=0 .$ Thus, the coordinates of the centre are $(1,-1)$.
Let $r$ be the radius of the circle.
$$
\begin{array}{l}
\pi r^{2}=154 \\
\Rightarrow \frac{22}{7} r^{2}=154 \Rightarrow r=7
\end{array}
$$
Hence, the equation of required circle is
$$
\begin{array}{l}
(x-1)^{2}+(y+1)^{2}=7^{2} \\
\Rightarrow x^{2}+y^{2}-2 x+2 y-47=0
\end{array}
$$
Hence, the correct answer is option: 'C'.