Question

Equation to the sphere which passes through the circle $x^2+y^2+z^2=5,x+2y+3z=3$ and touch the plane $4x+3y-$ $15=0$

• A. $x^2+y^2+z^2+2x+4y+6z+11=0$
• B. $x^2+y^2+z^2-2x-4y+6z-11=0$
• C. $x^2+y^2+z^2-2x-4y-6z+11=0$
• D. $x^2+y^2+z^2+2x+4y+6z-11=0$

Answer 1

Answer: (D)

$x^2+y^2+z^2+2x+4y+6z-11=0$

Equation to the sphere passes through the circle $x^2+y^2+z^2=5,x+2y+3z=3$ is given by,
$x^2+y^2+z^2-5+\lambda (x+2y+3z-3)=0$
$\Rightarrow x^2+y^2+z^2+\lambda x+2\lambda y+3\lambda z-(3\lambda +5)=0$
Given plane $4x+3y-15=0$ touches this sphere,
$\left|\frac{4\left(\frac{-\lambda }{2}\right)+3(-\lambda )-15}{5}\right|=\sqrt{{\left(\frac{\lambda }{2}\right)}^2+{\lambda }^2+{(\frac{3\lambda }{2}+3\lambda +5)}^2}$
$\Rightarrow 5{\lambda }^2-6\lambda -8=0\Rightarrow \lambda =2,-4/5$
Hence, the sphere is $x^2+y^2+z^2+2x+4y+6z-11=0$
There will be one more sphere corresponding to another value of $\lambda$.