Question

The equation of the circumcircle of the triangle formed by the lines $x=0,y=0,2x+3y=5$ is

• A. $x^2+y^2+2x+3y-5=0$
• B. $6(x^2+y^2)-5(3x+2y)=0$
• C. $x^2+y^2-2x-3y+5=0$
• D. $6(x^2+y^2)+5(3x+2y)=0$

Answer 1

The correct option is B $6(x^2+y^2)-5(3x+2y)=0$

So, the vertices are;-
$A(0,0),B(2.5,0)\&C(0,\frac{5}{4})$
Perpendicular bisector of AB
$x=\frac{5}{4})$
Perpendicular bisector of AC
$y=\frac{5}{6})$
So the two bisectors meet at
$0(\frac{5}{4},\frac{5}{6})$ Hence it is center
Radius$=A0=\sqrt{\frac{25}{16}+\frac{25}{36}}\simeq 1.5=\frac{3}{2}$
$\Rightarrow {(x-\frac{5}{4})}^2+{(y-\frac{5}{6})}^2={\left(\frac{3}{2}\right)}^2=\frac{25}{16}+\frac{25}{36}$
$\Rightarrow x^2+y^2-\frac{5}{2}x-\frac{5}{3}=0$
$\Rightarrow 6(x^2+y^2)-5(3x+2y)=0$