Question

Find the equation of the circle whose diameters are along the lines $2x-3y+12=0$ and $x+4y-5=0$ and whose area is $154sq$. units.

Answer 1

$\begin{array}{c}Letcentre\left(g,t\right)andradiusr.\\ thenaccordingtotheproblem,\\ 2h-3k+12=0---\left(i\right)\\ h+4k-5=0---\left(ii\right)\\ subtracting\left(i\right)-\left(ii\right)weget\\ \left(2h-3k+12\right)-\left(2h+8k-10\right)=0\\ -3k-8k+12+10=0\\ -11k=-22\\ \therefore k=2\\ h=-3\\ \therefore centre=\left(-3,2\right)\\ Now,\\ Area=154\\ \pi r^2=154\\ \frac{22}{7}.r^2=154\\ r^2=49\\ \therefore r=7cm\\ Hence,\\ eq{u}^{n}ofcircle:\\ {\left(x+3\right)}^2+{\left(y-2\right)}^2=7^2\\ {\left(x+3\right)}^2+{\left(y-2\right)}^2=49\end{array}$

Which is the required answer.