Q.If the lines $2x-3y=5$ and $3x-4y=7$ are the diameters of a circle of area $154$ sq. units, then find the equation of the circle. (A) $x^2+y^2-2x+2y=40$ (B) $x^2+y^2-2x-2y=47$ (C) $x^2+y^2-2x+2y=47$ (D) $x^2+y^2-2x-2y=40$

The centre is the intersection of $2x-3y=5$ and $3x-4y=7$. From the first, $x=\dfrac{5+3y}{2}$; substituting into the second: $3\left(\dfrac{5+3y}{2}\right)-4y=7 \Rightarrow 15+9y-8y=14 \Rightarrow y=-1$, so $x=1$. Centre $=(1,-1)$. Area $=154=\pi r^2$; using $\pi\approx\dfrac{22}{7}$, $r^2=154\times\dfrac{7}{22}=49$. Centre-radius form: $(x-1)^2+(y+1)^2=49$, expanding to $x^2+y^2-2x+2y+2-49=0$, i.e. $x^2+y^2-2x+2y-47=0$, i.e. $x^2+y^2-2x+2y=47$.

Solve the two diameter-lines simultaneously for the centre; use Area $=\pi r^2$ (with $\pi\approx22/7$) to get $r^2$; apply the centre-radius form.

• Using $\pi\approx3.14$ instead of $22/7$, which would not give the clean $r^2=49$ the options expect
• Sign slip moving the constant across to match the option's format $=47$ rather than $-47=0$