Question

The lines $2x-3y=5$ and $3x-4y=7$ are the diameters of a circle of area $154$ square units. An equation of this circle is $(\pi =22/7)$

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

Answer 1

Answer: (C)

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

The centre of the circle is the point of intersection of the given diameters
$2x-3y=5$ and $3x-4y=7,$ which is $(1,-1)$
and if $r$ is the radius of the circle,then
$\pi r^2=154\Rightarrow r^2=154\times \frac{7}{22}\Rightarrow r=7.$
Hence equation of the required
circle is $(x-1{)}^2+(y+1{)}^2=7^2\Rightarrow x^2+y^2-2x+2y=47$