WAEC 2013 · Paper 2 · Q4

Coordinate geometry & circles

(a)
The line $2y = x + 3$ meets the circle $x^2 + y^2 - 2x + 6y - 15 = 0$ at points $M$ and $N$ , where $N$ is in the first quadrant. Find the coordinates of $M$ and $N$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

From the line, $x = 2y - 3$ .
Substitute into the circle.
$(2y - 3)^2 + y^2 - 2(2y - 3) + 6y - 15 = 0$ .
$4y^2 - 12y + 9 + y^2 - 4y + 6 + 6y - 15 = 0$ , so $5y^2 - 10y = 0$ .
$5y(y - 2) = 0$ : $y = 0$ gives $x = -3$ , and $y = 2$ gives $x = 1$ .
$N$ is in the first quadrant, so $N(1, 2)$ and $M(-3, 0)$ .

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