Question

The line $2x+3y=1$ intersects the circle $x^2+$ $y^2=4$ at A and B. If the equation of the circle on AB as diameter is $x^2+y^2+2gx+2fy+c=0$, then $c=$

• A. $-50$
• B. $-\frac{54}{13}$
• C. $-\frac{54}{19}$
• D. $-\frac{50}{13}$

Answer 1

The correct option is D $-\frac{50}{13}$

$P$ is a foot of perpendicular from $(0,0)$ to the line $2x+3y=1$, then
$\frac{x-0}{2}=\frac{y-0}{3}=\frac{-(-1)}{2^2+3^2}=\frac{1}{13}$
$x=\frac{2}{13},y=\frac{3}{13}$
So, $P(\frac{2}{13},\frac{3}{13})$ is a centre of the circle with $AB$ as diameter.
So, radius of that circle is
$r=AP=\sqrt{4-{\left(\frac{1}{\sqrt{13}}\right)}^2}$
$=\sqrt{\frac{51}{13}}$
$\therefore$ Equation of the circle is
${(x-\frac{2}{13})}^2+{(y-\frac{3}{13})}^2=\frac{51}{13}$
$\Rightarrow x^2+y^2-\frac{4x}{13}-\frac{6y}{13}+\frac{1}{13}=\frac{51}{13}$
$\Rightarrow x^2+y^2-\frac{4x}{13}-\frac{6y}{13}-\frac{50}{13}=0$
$\therefore c=-\frac{50}{13}$