Question

Find the equation circle if the equations of two diameters are $2x+y=6$ and $3x+2y=4$. When radius of circle is $9$.

Answer 1

Intersection of equation of diameter gives centre

$3x+2y=4$

$2x+y=6\to \times 2$

$3x+2y=4$

$4x+2y=12$

$-x=-8$

$x=8$

$2x+y=6$

$16+y=6$

$y=-10$

Centre $C(8,-10)$

$Radius=r=9$

Equation of circle $C=(8,-10),r=9$

$(x-8{)}^2+(y+10{)}^2=(9{)}^2$

$x^2-16x+164+y^2+20y+100=81$

$x^2+y^2-16x+20y+83=0$.