Question

The point of contact of two spheres $x^2+y^2+z^2=25$,$x^2+y^2+z^2-18x-24y-40z+225=0$

• A. $(5,0,0)$
• B. $(\frac{9}{5},\frac{12}{5},-3)$
• C. $(\frac{9}{5},\frac{12}{5},4)$
• D. $(0,3,4)$

Answer 1

Answer: (C)

$(\frac{9}{5},\frac{12}{5},4)$

Given spheres are $x^2+y^2+z^2=25$ and $x^2+y^2+z^2-18x-24y-40z+225=0$

Let $(x_1,y_1,z_1)$ be a point that belong to both spheres, then it satisfies both equations and any linear combination of them.

In particular, the linear combination is

$x^2+y^2+z^2-(x^2+y^2+z^2-18x-24y-40z)=25-(-225)$

$⟹18x+24y+40z=225+25$

$⟹9x+12y+20z=125$ ......... $\left(i\right)$

So, the point belong to the plane $9x+12y+20z=125$. i.e. the point $(x_1,y_1,z_1)$ satisfy the equation of plane obtained.

Now, only the point $(\frac{9}{5},\frac{12}{5},4)$ satisfy $\left(i\right)$

i.e. $9\times \frac{9}{5}+12\times \frac{12}{5}+20\times 4=\frac{225}{5}+80=125$

Hence, $(\frac{9}{5},\frac{12}{5},4)$ is the point of contact.