Question

Equation of the sphere which touches the plane $3x+2y-z+2=0$ at the point $(1,-2,1)$ and cuts orthogonally the sphere $x^2+y^2+z^2-4x+6y+4=0$

• A. ${(x+\frac{7}{2})}^2+(y+5{)}^2+{(z-\frac{5}{2})}^2={\left(\frac{3\sqrt{14}}{2}\right)}^2$
• B. ${(x-\frac{7}{2})}^2+(y-5{)}^2+{(z+\frac{5}{2})}^2=(3\sqrt{14}{)}^2$
• C. ${(x-\frac{7}{2})}^2+(y+10{)}^2+(z+5{)}^2={\left(\frac{3}{2}\right)}^2$
• D. $(x+7{)}^2+(y-10{)}^2+(z-5{)}^2=(3{)}^2$

Answer 1

The correct option is A ${(x+\frac{7}{2})}^2+(y+5{)}^2+{(z-\frac{5}{2})}^2={\left(\frac{3\sqrt{14}}{2}\right)}^2$

The given plane is $3x+2y-z+2=0$ ...$\left(1\right)$

and the sphere is $x^2+y^2+z^2-4x+6y+4=0$ ...$\left(2\right)$ ......$\because$ the required sphere touches the plane $\left(1\right)$ at $P(1,-2,1)$

$\therefore$ its centre lies on the normal t the plane(1) through $P(1,-2,1)$ are $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z-1}{-1}$

Any point on this line is $C(3r+1,2r-2,-r+1)$

Let this be the centre of the required sphere.

The radius of required sphere $=CP$

$=\sqrt{{(3r+1-1)}^2+{(2r-2+2)}^2+{(-r+1-1)}^2}=\sqrt{9r^2+4r^2{r}^2}=r\sqrt{14}$

$\because$ the required sphere cuts $\left(2\right)$ orthogonally

$\therefore$ squares of the distance between the centres$=$sum of squares of their radii ...$\left(3\right)$

Now the centre of the sphere $\left(2\right)$ is $(2,-3,0)$ and radius is $\sqrt{4+9-4}=3$

Also the centreand radius of required sphere are $(3r+1,2r-2,-r+1)$ and $\sqrt{14}r$

$\therefore$From (3), ${(3r+1-2)}^2+{(2r-2+3)}^2+{(-r+1-0)}^2=9+14r^2\Rightarrow r=-\frac{3}{2}$

$\therefore$ coordinates of $c$ are $(-\frac{7}{2},-5,\frac{3}{2})$

and the radius of the required sphere is $\sqrt{14}r=\sqrt{14}\left(\frac{3}{2}\right)=\frac{3\sqrt{14}}{2}$

$\therefore$ the required sphere is ${(x+\frac{7}{2})}^2+{(y+5)}^2+{(z-\frac{5}{2})}^2={\left(\frac{3\sqrt{14}}{2}\right)}^2$