Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# What is the diameter of the sphere ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x+6y-8z-7=0$?A. 4 unitsB. 5 unitsC. 6 unitsD. 12 units

Last updated date: 22nd Jun 2024
Total views: 404.1k
Views today: 9.04k
Verified
404.1k+ views
Hint: We here have been given the equation of the sphere as ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x+6y-8z-7=0$ and we have to find its diameter. For this, we will equate this equation with the general equation of the sphere, i.e. ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2fx+2gy+2hz+c=0$ and get the values of f, g, h and c from the given equation. Then we will use the formula for the radius of a sphere given as $r=\sqrt{{{f}^{2}}+{{g}^{2}}+{{h}^{2}}-c}$ and hence obtain the radius of the sphere. Then we will use the fact that the diameter of a sphere is twice its radius and hence find the value of the diameter.

Here, we have been given the equation of a sphere as ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x+6y-8z-7=0$ and we have to find its diameter.
Now, we can see that this equation of the sphere is in the form of ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2fx+2gy+2hz+c=0$ which is the general equation of a sphere.
We also know that the radius of a sphere with the equation ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2fx+2gy+2hz+c=0$ is given as:
$r=\sqrt{{{f}^{2}}+{{g}^{2}}+{{h}^{2}}-c}$
Where r is the radius of the sphere.
Now, if we equate the given equation of the sphere, i.e. ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x+6y-8z-7=0$ with the general equation of the sphere ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2fx+2gy+2hz+c=0$, we will get:
2f=-4
2g=6
2h=-8
c=-7
Hence, on solving these, we get the value of f, g and h as:
\begin{align} & f=\dfrac{-4}{2}=-2 \\ & g=\dfrac{6}{2}=3 \\ & h=\dfrac{-8}{2}=-4 \\ \end{align}
Now, if we put these values of f, g, h and c in the formula for radius, we will get the radius of the required sphere.
Thus, putting the values of f, g, h and c in the formula for radius, we get:
\begin{align} & r=\sqrt{{{f}^{2}}+{{g}^{2}}+{{h}^{2}}-c} \\ & \Rightarrow r=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 3 \right)}^{2}}+{{\left( -4 \right)}^{2}}-\left( -7 \right)} \\ \end{align}
Now, solving this, we get the radius as:
\begin{align} & r=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( 3 \right)}^{2}}+{{\left( -4 \right)}^{2}}-\left( -7 \right)} \\ & \Rightarrow r=\sqrt{4+9+16+7} \\ & \Rightarrow r=\sqrt{36} \\ & \Rightarrow r=6 \\ \end{align}
Thus, the radius of the given sphere is 6 units. But here, we need to find the diameter.
Now, we know that the diameter of any sphere is twice its radius.
Thus, we can say that:
$diameter=2\left( radius \right)$
Putting the value of radius in this, we get:
\begin{align} & diameter=2\left( radius \right) \\ & \Rightarrow diameter=2\left( 6 \right) \\ & \therefore diameter=12units \\ \end{align}
Thus, the diameter of the given sphere is 12 units.

So, the correct answer is “Option D”.

Note: We can also do this question by the following method:
Now, we have been given the equation of the sphere as ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x+6y-8z-7=0$.
Now, we know that the general equation of a sphere is given as:
${{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}+{{\left( z-\gamma \right)}^{2}}={{r}^{2}}$
Where $\left( \alpha ,\beta ,\gamma \right)$ is its centre and r is the radius.
Thus, we can change the given equation in this form and get our radius as:
\begin{align} & {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x+6y-8z-7=0 \\ & \Rightarrow \left( {{x}^{2}}-4x \right)+\left( {{y}^{2}}+6y \right)+\left( {{z}^{2}}-8y \right)-7=0 \\ \end{align}
Adding 4, 9 and 16 on the LHS and the RHS, we get:
\begin{align} & \left( {{x}^{2}}-4x \right)+\left( {{y}^{2}}+6y \right)+\left( {{z}^{2}}-8y \right)-7=0 \\ & \Rightarrow \left( {{x}^{2}}-4x \right)+\left( {{y}^{2}}+6y \right)+\left( {{z}^{2}}-8y \right)-7+4+9+16=4+9+16 \\ & \Rightarrow \left( {{x}^{2}}-4x+4 \right)+\left( {{y}^{2}}+6y+9 \right)+\left( {{z}^{2}}-8y+16 \right)-7=29 \\ & \Rightarrow {{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}+{{\left( z-4 \right)}^{2}}=29+7 \\ & \Rightarrow {{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}+{{\left( z-4 \right)}^{2}}=36 \\ & \Rightarrow {{\left( x-2 \right)}^{2}}+{{\left( y+3 \right)}^{2}}+{{\left( z-4 \right)}^{2}}={{\left( 6 \right)}^{2}} \\ \end{align}
Thus, the radius of the given sphere is 6.
Hence, the diameter is given as:
\begin{align} & diameter=2\left( radius \right) \\ & \Rightarrow diameter=2\left( 6 \right) \\ & \therefore diameter=12units \\ \end{align}