Question

How do you find an equation for each sphere that passes through the point \[(5,1,4)\] and is tangent to all three coordinate planes?

Answer 1

Hint: Before solving this question, we should know about the equation of the sphere in the coordinate plane, and also it should be known that all the coordinate planes lie at equal distance from each other. After knowing all these, it will be easy for us to solve the question.

Complete step-by-step solution:
The \[3D\] representation of a circle is known as a sphere. The coordinates of the sphere tell us that where the sphere lies in the coordinate system
In the language of mathematics, a sphere can also be defined as the set of all the points which are equidistant from a given point in a three-dimensional plane.
Now in this question, we will learn to find the equation of each sphere that passes through a given point and is tangent to all the three planes in the Cartesian system.
The general equation of sphere is given as-
\[{{(x-{{x}_{0}})}^{2}}+{{(y-{{y}_{0}})}^{2}}+{{(z-{{z}_{0}})}^{2}}={{r}^{2}}\]
The coordinates of the center of the sphere are \[({{x}_{0}},{{y}_{0}},{{z}_{0}})\] but as the coordinate planes are equidistant from each other and also the radius of the sphere is equal, so the coordinates of the center of the sphere will be \[(r,r,r)\]
So from this the following equation is obtained.
\[{{(x-r)}^{2}}+{{(y-r)}^{2}}+{{(z-r)}^{2}}={{r}^{2}}\]…….eq(1)
Now we will solve this equation further by putting the values of x,y, and z in the above equation which is as follows.
\[{{(5-r)}^{2}}+{{(1-r)}^{2}}+{{(4-r)}^{2}}={{r}^{2}}\]
Now we will solve this equation further by opening the brackets.
\[\Rightarrow {{(5)}^{2}}+{{r}^{2}}-10r+{{(1)}^{2}}+{{r}^{2}}-2r+{{(4)}^{2}}+{{r}^{2}}-8r={{r}^{2}}\]
\[\Rightarrow 25+1+16+3{{r}^{2}}-20r={{r}^{2}}\]
\[\Rightarrow {{r}^{2}}-10r+21=0\]
Now we will solve the above equation by splitting the middle term method by which the value of r will be obtained which is as follows.
\[\begin{align}
  & {{r}^{2}}-(7+3)r+21=0 \\
 & \Rightarrow {{r}^{2}}-7r-3r+21=0 \\
\end{align}\]
\[\Rightarrow r(r-7)-3(r-7)=0\]
\[\Rightarrow (r-7)(r-3)=0\]
So now two values of radius r are obtained which is given below
\[\begin{align}
  & r=7 \\
 & r=3 \\
\end{align}\]
Now we will put these values one by one in eq(1) by which the required equation will be obtained.
First, we will put the value of \[r=7\] in eq(1), then we get
\[{{(x-7)}^{2}}+{{(y-7)}^{2}}+{{(z-7)}^{2}}={{7}^{2}}\]
After that we will put the value of \[r=3\] in eq(1), then we get
\[{{(x-3)}^{2}}+{{(y-3)}^{2}}+{{(z-3)}^{2}}={{3}^{2}}\]

Note: A circle is a \[2\] dimensional figure whereas a sphere is a \[3\] dimensional figure. The origin of the coordinate system is the point where all the coordinate planes intersect each other. There is an angle named dihedral angle which is the angle between two spheres when they intersect each other.