If the lines 2x+3y+1=0 and 3x-y-4=0 lie along diameters of a circle of circumference 10, t…

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:

Accepted Answer

Two diameters are along

$2x+3y+1=0$ and $3x-y-4=0$

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

circumference $$ = 2\pi r = 10\pi $$

$$\therefore$$ $r=5$.

Required circle is, $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {5^2}$$

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

Solutions · 1