Question

The centre of the circle $x^2+y^2+z^2-3x+4y-2z-5=0$ and $5x-2y+4z+7=0$ is

• A. $(\frac{3}{2},-2,1)$
• B. $(1,1,1)$
• C. $(-1,-1,-1)$
• D. $(0,0,0)$

Answer 1

Answer: (C)

$(-1,-1,-1)$

Find the foot of perp. from centre $(\frac{3}{2},-2,1)$ to the plane. Any line through centre, perpendicular to plane i.e., parallel to normal is
$\frac{x-3/2}{5}=\frac{y+2}{-2}=\frac{z-1}{4}=r$
Any point on the line is $(5r+3/2,-2r-2,4r+1)$. It lies on given plane.
$\therefore 5(5r+\frac{3}{2})-2(-2r-2)+4(4r+1)+7=0$
$r(25+4+16)+(\frac{15}{2}+4+4+7)=0$
$\therefore 45r+\frac{45}{2}=0$

$\therefore r=-\frac{1}{2}$
$\therefore$ foot of perpendicular or centre of circle is $(-1,-1,-1)$.