Question

A circle with center $(2,3,0)$ and radius $1$, is drawn in the plane $z=0$. Equation of the sphere which passes through this circle and through the point $(1,1,1)$ is

• A. $x^2+y^2+z^2-4x-6y+7=0$
• B. $x^2+y^2+z^2-4x-6y+2z+5=0$
• C. $x^2+y^2+z^2+4x+6y-13=0$
• D. $x^2+y^2+z^2-4x-6y-5z+12=0$

Answer 1

Answer: (D)

$x^2+y^2+z^2-4x-6y-5z+12=0$

Equation of the given circle is

${(x-2)}^2+{(y-3)}^2+z^2=1$

And the plane by $z=0$

Equation of any sphere through the circle is

${(x-2)}^2+{(y-3)}^2+z^2-1+\lambda z=0$

$\Rightarrow x^2+y^2+z^2-4x-6y+\lambda z+12=0$

Since, it passes through $(1,1,1)$

$1+1+1-4-6+\lambda +12=0$

$\Rightarrow \lambda =-5$

And the required equation is

$x^2+y^2+z^2-4x-6y-5z+12=0$

Hence, option D.