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Question: What is the diameter of the sphere \({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x+6y-8z-7=0\)? A. 4 units B....

What is the diameter of the sphere x2+y2+z2−4x+6y−8z−7=0{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x+6y-8z-7=0?
A. 4 units
B. 5 units
C. 6 units
D. 12 units

Explanation

Solution

We here have been given the equation of the sphere as x2+y2+z2−4x+6y−8z−7=0{{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. x2+y2+z2+2fx+2gy+2hz+c=0{{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=f2+g2+h2−cr=\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.

Complete step by step answer:
Here, we have been given the equation of a sphere as x2+y2+z2−4x+6y−8z−7=0{{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 x2+y2+z2+2fx+2gy+2hz+c=0{{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 x2+y2+z2+2fx+2gy+2hz+c=0{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2fx+2gy+2hz+c=0 is given as:
r=f2+g2+h2−cr=\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. x2+y2+z2−4x+6y−8z−7=0{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x+6y-8z-7=0 with the general equation of the sphere x2+y2+z2+2fx+2gy+2hz+c=0{{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:
f=−42=−2 g=62=3 h=−82=−4 \begin{aligned} & f=\dfrac{-4}{2}=-2 \\\ & g=\dfrac{6}{2}=3 \\\ & h=\dfrac{-8}{2}=-4 \\\ \end{aligned}
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:
r=f2+g2+h2−c ⇒r=(−2)2+(3)2+(−4)2−(−7) \begin{aligned} & 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{aligned}
Now, solving this, we get the radius as:
r=(−2)2+(3)2+(−4)2−(−7) ⇒r=4+9+16+7 ⇒r=36 ⇒r=6 \begin{aligned} & 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{aligned}
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(radius)diameter=2\left( radius \right)
Putting the value of radius in this, we get:
diameter=2(radius) ⇒diameter=2(6) ∴diameter=12units \begin{aligned} & diameter=2\left( radius \right) \\\ & \Rightarrow diameter=2\left( 6 \right) \\\ & \therefore diameter=12units \\\ \end{aligned}
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 x2+y2+z2−4x+6y−8z−7=0{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4x+6y-8z-7=0.
Now, we know that the general equation of a sphere is given as:
(x−α)2+(y−β)2+(z−γ)2=r2{{\left( x-\alpha \right)}^{2}}+{{\left( y-\beta \right)}^{2}}+{{\left( z-\gamma \right)}^{2}}={{r}^{2}}
Where (α,β,γ)\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:
x2+y2+z2−4x+6y−8z−7=0 ⇒(x2−4x)+(y2+6y)+(z2−8y)−7=0 \begin{aligned} & {{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{aligned}
Adding 4, 9 and 16 on the LHS and the RHS, we get:
(x2−4x)+(y2+6y)+(z2−8y)−7=0 ⇒(x2−4x)+(y2+6y)+(z2−8y)−7+4+9+16=4+9+16 ⇒(x2−4x+4)+(y2+6y+9)+(z2−8y+16)−7=29 ⇒(x−2)2+(y+3)2+(z−4)2=29+7 ⇒(x−2)2+(y+3)2+(z−4)2=36 ⇒(x−2)2+(y+3)2+(z−4)2=(6)2 \begin{aligned} & \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{aligned}
Thus, the radius of the given sphere is 6.
Hence, the diameter is given as:
diameter=2(radius) ⇒diameter=2(6) ∴diameter=12units \begin{aligned} & diameter=2\left( radius \right) \\\ & \Rightarrow diameter=2\left( 6 \right) \\\ & \therefore diameter=12units \\\ \end{aligned}