Question

The line $2x+3y=1$ cuts the circle $x^2+y^2=4$ in $P$ and $Q$. Then the equation of the circle on $PQ$ as diameter is

• A. $13(x^2+y^2)-4x-6y+50=0$
• B. $13(x^2+y^2)-6y-50=0$
• C. $13(x^2+y^2)-4x-6y-50=0$
• D. $13(x^2+y^2)-4x-50=0$

Answer 1

The correct option is C $13(x^2+y^2)-4x-6y-50=0$

s: $x^2+y^2=4$
$radius=c_1=z$
foot of perpendicutor from centre $(0,0)to2x+3y=1$ line is point PQ.
$\therefore \frac{h-0}{2}=\frac{k-0}{3}=-1\frac{(-1)}{2^2+3^2}=\frac{1}{13}$
$h=\frac{+2}{13}k=\frac{+3}{13}$
$\therefore PM=MQ=\sqrt{c{p}^2-c{m}^2}$
$=\sqrt{4-{\left(\sqrt{\frac{13}{13\times 13}}\right)}^2}$
$=\sqrt{4-\frac{1}{13}}$
$=\sqrt{\frac{51}{13}}$
$\therefore e{q}^n$of circle having PQ of diameter and centre $(\frac{2}{13},\frac{3}{13})$
$\therefore {(x-\frac{2}{13})}^2+{(y-\frac{3}{13})}^2=\frac{51}{13}$
$x^2+y^2-\frac{4x}{13}-\frac{6y}{13}-\frac{50}{13}=0$
$\Rightarrow 13(x^2+y^2)-4x-6y-50=0$