Question

The straight line $x+y-1=0$ meets the circle $x^2+y^2-6x-8y=0$ at A and B. Then the equation of the circle of which AB is a diameter is

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

Answer 1

Answer: (A)

$x^2+y^2-2y-6=0$

$x^2+y^2-6x-8y+\lambda (x+y-1)=0$
$\therefore$ Centre $=\left(\begin{array}{c}3-\frac{\lambda }{2},4-\frac{\lambda }{2}\end{array}\right)$ lies on $x+y-1=0$
So, the centre satisfy the equation $x+y-1=0$
$\therefore \left(\begin{array}{c}3-\frac{\lambda }{2}\end{array}\right)+\left(\begin{array}{c}4-\frac{\lambda }{2}\end{array}\right)-1=0$
$\Rightarrow 6-\lambda =0\Rightarrow \lambda =6$
Hence, required equation is given by
$x^2+y^2-6x-8y+6x+6y-6=0$
$\Rightarrow x^2+y^2-2y-6=0$