NECO 2023 · Paper 2 · Q9

Coordinate geometry & circles

(a)
Find the centre and radius of the circle $3x^2 + 3y^2 + 12x - 6y - 45 = 0$ .
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Centre $(-2, 1)$ , radius $2\sqrt5$
(b)
Find the equation of the tangent to the circle at $(2, 3)$ (give $y$ in terms of $x$ ).
(c)
Find the value of the angle $\theta$ in the parametric coordinates of the point $(2, 3)$ (degrees, 2 d.p.).

Worked solution (try it first)

(a)

Divide by 3: $x^2 + y^2 + 4x - 2y - 15 = 0$ , so $g = 2$ and $f = -1$ .
The centre is $(-g, -f) = (-2, 1)$ and $r = \sqrt{4 + 1 + 15}$
$= \sqrt{20}$
$= 2\sqrt5$ .

(b)

The radius to $(2, 3)$ has gradient $\dfrac{3 - 1}{2 + 2} = \dfrac12$ .
The tangent is perpendicular to it, with gradient $-2$ : $y - 3 = -2(x - 2)$ , so $y = -2x + 7$ .

(c)

On the circle, $x = -2 + 2\sqrt5\cos\theta$ and $y = 1 + 2\sqrt5\sin\theta$ .
At $(2, 3)$ : $\cos\theta = \dfrac{4}{2\sqrt5}$
$= \dfrac{2}{\sqrt5}$ and $\sin\theta = \dfrac{2}{2\sqrt5}$
$= \dfrac{1}{\sqrt5}$ .
So $\tan\theta = \frac12$ with both positive: $\theta \approx 26.57^\circ$ .

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