NECO 2023 · Paper 2 · Q2

(a)
Find the equation of the circle whose centre is $(4, -5)$ and which passes through $(-3, 2)$ .
Show the answer
$x^2 + y^2 - 8x + 10y - 57 = 0$
(b)
Simplify $\left(\dfrac{x^{-\frac23}}{\sqrt[3]{x^2}} \div \dfrac{1}{x^2}\right)^{-3}$ .

Worked solution (try it first)

(a)

The radius is the distance from the centre $(4, -5)$ to the point $(-3, 2)$ : $r^2 = (4 - (-3))^2 + (-5 - 2)^2$ .
So $r^2 = 7^2 + (-7)^2 = 49 + 49 = 98$ .
The circle with centre $(a, b)$ and radius $r$ is $(x - a)^2 + (y - b)^2 = r^2$ .
Here: $(x - 4)^2 + (y + 5)^2 = 98$ .
Expand the brackets: $x^2 - 8x + 16 + y^2 + 10y + 25 = 98$ .
Collect everything on one side: $x^2 + y^2 - 8x + 10y - 57 = 0$ .

(b)

Write the root as a power: $\sqrt[3]{x^2} = x^{\frac23}$ .
Dividing, $x^{-\frac23} \div x^{\frac23} = x^{-\frac23 - \frac23}$
$= x^{-\frac43}$ .
Dividing by $\frac{1}{x^2}$ is multiplying by $x^2$ : $x^{-\frac43} \times x^2 = x^{-\frac43 + 2}$
$= x^{\frac23}$ .
Raise to the power $-3$ by multiplying the indices: $\left(x^{\frac23}\right)^{-3} = x^{-2}$ , that is $\dfrac{1}{x^2}$ .

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