Question

The lines $2x-3y-5=0$ and $3x-4y=7$ are diameters of a circle of area 154 sq units, then the equation of the circle is.( Use $\pi =\frac{22}{7}$)

• A. $x^2+y^2+2x-2y-62=0$
• B. $x^2+y^2+2x-2y-47=0$
• C. $x^2+y^2-2x+2y-47=0$
• D. $x^2+y^2-2x-2y-62=0$

Answer 1

Answer: (C)

$x^2+y^2-2x+2y-47=0$

The centre of the required circle lies at the intersection of $2x-3y-5=0$ and $3x-4y-7=0$.

Thus, the coordinates of the centre are $(1,-1)$.

Let $r$ be the radius of the circle.

$\pi r^2=154$

$\Rightarrow \frac{22}{7}{r}^2=154\Rightarrow r=7$

Hence, the equation of required circle is $(x-1{)}^2+(y+1{)}^2=7^2$

$\Rightarrow x^2+y^2-2x+2y-47=0$