Question

A point inside the circle $x^2+y^2+3x-3y+2=0$ is

• A. $(-1,3)$
• B. $(-2,1)$
• C. $(2,1)$
• D. $(-3,2)$

Answer 1

The correct option is B $(-2,1)$

$x^2+y^2+3x-3y+2=0$

$(x+\frac{3}{2}{)}^2+(y-\frac{3}{2}{)}^2+2=(\frac{3}{2}{)}^2+(+\frac{3}{2}{)}^2$

$(x+\frac{3}{2}{)}^2+(y-\frac{3}{2}{)}^2=\frac{9}{2}-2=\frac{5}{2}=(\sqrt{\frac{5}{2}}{)}^2$

center o $(\frac{-3}{2},\frac{3}{2})$ radius $=\sqrt{\frac{5}{2}}$

$\left(A\right)(-1,3)\to OA=\sqrt{(\frac{-3}{2}+1{)}^2+(\frac{3}{2}-3{)}^2}$

$=\sqrt{(1/2{)}^2+(-3/2{)}^2}=\frac{\sqrt{10}}{2}=\sqrt{\frac{5}{2}}$

$\because$ OA = Radius

$\therefore$ A Lies on the circle

$\left(B\right).(-2,1)\to OB=\sqrt{(-2+3/2{)}^2+(1-3/2{)}^2}$

$OB=\sqrt{\frac{1}{4}+\frac{1}{4}}=\sqrt{1/2}$

$\because OB<$ Radius

$\therefore$ B lies inside circle