Question 8
Point $A$ is the centre of a sphere and point $B$ lies on its surface. The booklet’s diagram labels them $A(1,0,-1)$ and $B(1,-3,-5)$.
The equation of the sphere is
That’s it. Step through the hints below to see why. In 2023, 42.3% of Queensland students chose (A).
Not quite. The signs say the centre is $(-1,0,1)$. For centre $(1,0,-1)$ you need $(x-1)^2$ and $(z+1)^2$. In 2023, 6.4% of Queensland students chose (B).
Not quite. The centre is right, but the two $1$s from expanding $(x-1)^2$ and $(z+1)^2$ still need to move across: $25-1-1=23$. In 2023, 33.4% of Queensland students chose (C).
Not quite. Both the signs and the constant are off. The centre would be $(-1,0,1)$, and the expanded constants haven’t moved across. In 2023, 17.5% of Queensland students chose (D).
This matches option (A).
The signs of the $x$ and $z$ terms give the centre. Check them against $A(1,0,-1)$.
After expanding, the constants $1$ and $1$ move to the right-hand side, so $25$ becomes $23$.
The radius is the distance from the centre $A$ to the point $B$ on the surface.
$$r=\sqrt{(1-1)^2+(-3-0)^2+(-5+1)^2}=\sqrt{0+9+16}=5$$
A sphere with centre $(a,b,c)$ and radius $r$ is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$.
$$(x-1)^2+y^2+(z+1)^2=25$$
Expand and move the constants across.
$$x^2-2x+1+y^2+z^2+2z+1=25\ \Rightarrow\ x^2-2x+y^2+z^2+2z=23$$
The radius is $|\overrightarrow{AB}|=\sqrt{0^2+3^2+4^2}=5$.
$$(x-1)^2+y^2+(z+1)^2=25\ \Rightarrow\ x^2-2x+y^2+z^2+2z=23$$
How Queensland students answered in 2023
From the QCAA Specialist Mathematics subject report for 2023. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2023 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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