Question

Find the equation of a sphere which passes through
origin and intercepts lengths a, b and c on the axes
respectively.
Or
Find the equation of a sphere passing through origin
and the points, $(a,0,0),(0,b,0)$ and $(0,0,c).$
Or
If the plane $x/a+y/b+z/c=1$ meets the co - ordinate
axes in points A, B, C and O be the origin. Find the
sphere OABC.

• A. $x^2+y^2+z^2-ax-by-cz=0.$
• B. $x^2+y^2+z^2+ax+by+cz=0.$
• C. $x^2+y^2+z^2-cx-by-az=0.$
• D. $x^2+y^2+z^2+cx+by+az=0.$

Answer 1

The correct option is A $x^2+y^2+z^2-ax-by-cz=0.$

Since the circle cuts intercepts of a and b on the x axis, y axis and z axis respectivley.

$\therefore$ the line joining the points $(a,0,0),(0,b,0),(0,0,c)$ will be the diameter of the circle

Let the radius of the circle be $r$

$\therefore r^2=(0.5a-a{)}^2+(0.5b-b)+(0.5c^2-c)$

Thus equation of the circle is as follows

$(x-h{)}^2+(y-k{)}^2+(z-l{)}^2=r^2$

$\therefore$ Equation is $x^2+y^2+z^2-ax-by-cz=0.$