Question

The lines 2x-3y $=5$ and 3x-4y $=7$ diameters of a circle having area as $154$ units. Then the equation of the circle is:

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

Answer 1

The correct option is D $x^2+y^2-2x+2y=47$

$2x-3y=5----\left(1\right)$
$3x-4y=7----\left(2\right)$
are diameter of circle so, point of intersection is centre of circle.
from (1) & (2), $x=1,y=-1$
$\therefore$ area of circle is $154$
$=\pi r^2=154$
$r^2=\frac{7}{22}\times 154$
$=\frac{7\times 14}{2}$
$r^2=49$
$r=7$
$\therefore$ Equation of circle with centre $(1,-1)$ and radius $7$ units
$(x-1{)}^2+(y+1{)}^2=7^2$
$\Rightarrow x^2+y^2-2x+2y-47=0$