Question

If the two lines $x+y=6$ and $x+2y=4$ are the diameters of the circle which passes through $(2,6)$, then its equation is

• A. $x^2+y^2-16x+4y+32=0$
• B. $x^2+y^2-16x+4y+23=0$
• C. $x^2+y^2-16x+4y-32=0$
• D. $x^2+y^2-16x+4y-23=0$

Answer 1

The correct option is C $x^2+y^2-16x+4y-32=0$

Given lines:
$x+y=6,x+2y=4$
Point of intersection of these lines is the centre of the circle
$C\equiv (8,-2)$

Radius is the distance between $(8,-2)$ and $(2,6)$
$r=\sqrt{36+64}=10$

So, the equation of required circle is
$(x-8{)}^2+(y+2{)}^2={10}^2$
$\therefore x^2+y^2-16x+4y-32=0$