Question

Find the equation of a sphere whose centre is (1,2,3) and touches the plane x+2y+3z = 0

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

We saw that we can write the equation of a sphere if we know the centre and radius of the sphere. We are given the centre. Now, if we can find the radius also, we can write the equation of the sphere. For that, we are given one more condition; the sphere touches the plane. This means that the perpendicular distance of centre from the plane will be equal to the radius.( Think about the equivalent case when a line touches a circle. In that case also, perpendicular distance of center to the line would be equal to radius of circle)

So, we will find the perpendicular distance of centre from the plane to get the radius. We also saw that the perpendicular distance of a point (x1, y1, z1) from ax+by+cz+d = 0 is given by

$\left|\frac{ax1+by1+cz1+d}{\sqrt{a^2+b^2+c^2}}\right|$ $Radius=\left|\frac{1+2\times 2+3\times 3}{\sqrt{1^2+2^2+3^2}}\right|=\sqrt{14}$ Now we can directly write the equation of sphere. $(x-1{)}^2+(y-2{)}^2+(z-3{)}^2=({\sqrt{14}}^2)\Rightarrow x^2+y^2+z^2-2x-4y-6z=0$