Q. The line $2x - 3y = 5$ and $3x - 4y = 7$ are the diameter of a circle of ares 154 sq unit. The 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) $x^{2} + y^{2} - 2x + 2y = 62$
Explanation
(c)
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 the radius is $r$, where $\pi r^{2} = 154$
$\Rightarrow \ r^{2} = 154 \times \frac{7}{22}\ \ \Rightarrow \ r = 7$
and hence, the required equation of the circle is
$(x - 1)^{2} + (y + 1)^{2} = 7^{2}$
$\Rightarrow \ x^{2} + y^{2} - 2x + 2y = 47$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.