Question

If $x-y+1=0$ meets the circles $x^2+y^2+y-1=0$ at $A,B$, then the equation of the circle with $AB$ as diameter

• A. $2(x^2+y^2)+3x-y+1=0$
• B. $2(x^2+y^2)+3x-y+2=0$
• C. $2(x^2+y^2)+3x-y+3=0$
• D. $x^2+y^2+3x-y+1=0$

Answer 1

The correct option is B $2(x^2+y^2)+3x-y+2=0$

$x^2+y^2+y-1=0$
radius $=\sqrt{\frac{1}{4}+1}=\sqrt{\frac{5}{2}}$
$x-y+1=0$
Foot of perpendicular from $C(0,-1/2)onx-y+1=0$ given mid-point of diameter AB i.e point E
$\therefore \frac{x-0}{1}=\frac{y{+}^1{/}_2}{-1}=-1\frac{{(}^3{/}_2)}{1^2+1^2}=\frac{-3}{4}$
$x{=}^{-3}{/}_{4}andy=\frac{3}{4}\frac{-1}{2}=\frac{1}{4}$
$E\left({}^{-3}{/}_4{,}^1{/}_4\right)$
So, radius of circle having AB of diameter is
$R=\sqrt{r^2-(CF{)}^2}=\sqrt{{}^5{/}_4-\frac{9}{8}}$
$=\sqrt{\frac{1}{8}}=\frac{1}{2\sqrt{2}}$
$\therefore e{q}^n$ of circle having AB of diameter,
$\therefore {\left(x{+}^3{/}_4\right)}^2+{\left(y{-}^1{/}_4\right)}^2={\left(\frac{1}{2\sqrt{2}}\right)}^2=\frac{1}{8}$
$\Rightarrow 2(x^2+y^2)+3x-y+2=0$