Question

lf the lines $2x+3y+1=0$ and $3x-y-4=0$ lie along diameters of a circle of circumference $10\pi$, then the equation of the circle is:

• A. $x^2+y^2-2x+2y-23=0$
• B. $x^2+y^2-2x-2y-23=0$
• C. $x^2+y^2+2x+2y-23=0$
• D. $x^2+y^2+2x-2y-23=0$

Answer 1

The correct option is A $x^2+y^2-2x+2y-23=0$

$2x+3y+1=0$ and $3x-y-4=0$ lie along diameter of circle.

So, point of intersection of $2x+3y+1=0$ and $3x-7-4=0$ is a centre of required circle.

$\therefore 11x-11=0.$

$x=1$ & $y=-1.$

So, equation of circle having radius $r=\frac{10}{2\pi }=5$ and centre $(1,-1)$ is given by $(x-1{)}^2+(y+1{)}^2=25=r^2$

$\Rightarrow x^2+y^2-2x+2y-23=0$