Question

If 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 $x^2+y^2-2x+2y-23=0$

Two diameters are along

$2x+3y+1=0$

$3x-y-4=0$

$2x+3y+1=0$ ..............$\left(1\right)$

$3x-y-4=0$ ...............$\left(2\right)\times 3$

Subtract $\left(2\right)-\left(1\right)$

$\Rightarrow x=1$ and $y=-1$

on solving we get center as $(1,-1)$

We know that circumference$=2\pi r$

$2\pi r=10\pi$

$2r=10$

$r=5$

Required circle is $(x-1{)}^2+(y+1{)}^2=52$

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