Question

Find the equation of the circle intercepted by the line $2x+3y=6$ between the co-ordinate axes as the diameter.

Answer 1

Given the line $2x+3y=6$

Intercepts by the line on coordinate axes will be:

Put $x=0$ we get $y=2$ intercept on y-axis

Put $y=0$ we get $x=3$ intercept on x-axis

Mid point of $AB$ will be $C(\frac{3}{2},1)$

radius $r=\sqrt{{(\frac{3}{2}-0)}^2+(1-2{)}^2}=\frac{\sqrt{13}}{2}$

Equation of the circle with center $C$ will be ${(x-\frac{3}{2})}^2+(y-1{)}^2=\frac{13}{4}$

$\Rightarrow x^2+y^2-3x-2y=0$

Therefore the equation of required circle will be $x^2+y^2-3x-2y=0$