WAEC 2008 · Paper 2 · Q10

(a)
The equation of a circle is $x^2 + y^2 - 4x + 2y + c = 0$ , where $c$ is a constant. If the radius of the circle is $2\sqrt3$ , find the value of $c$ .
(b)
$T$ is the tangent to the curve $y = x^2 + 6x - 4$ at $(1, 3)$ and $N$ is the normal to the curve $y = x^2 - 6x + 18$ at $(4, 10)$ . Find the coordinates of the point of intersection of $T$ and $N$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

For $T$ : $\frac{dy}{dx} = 2x + 6$ , which is $8$ at $x = 1$ .
Tangent through $(1, 3)$ with gradient 8: $y - 3 = 8(x - 1)$ , so $y = 8x - 5$ .
For $N$ : $\frac{dy}{dx} = 2x - 6$ , which is $2$ at $x = 4$ .
The normal has gradient $-\frac12$ .
Normal through $(4, 10)$ : $y - 10 = -\frac12(x - 4)$ , so $x + 2y = 24$ .
Substitute $y = 8x - 5$ : $x + 16x - 10 = 24$ , so $17x = 34$ and $x = 2$ .
Then $y = 8(2) - 5 = 11$ .
$T$ and $N$ meet at $(2, 11)$ .