The circle c has centre P(-2, –1) and passes through the point Q(3, 1) - Leaving Cert Mathematics - Question 1 - 2010

The radius of the circle, $r$, is the distance from the center P(-2, -1) to the point Q(3, 1) as calculated earlier:

$r = ext{sqrt}(29)$

To write down the equation of the circle in standard form, use the formula:

$(x - h)^2 + (y - k)^2 = r^2$

where (h, k) is the center. Substituting the values, we have:

$(x + 2)^2 + (y + 1)^2 = 29$

To verify that QR is a tangent to the circle, we need to calculate the slopes of PQ and QR.

1. Slope of PQ: The slope formula is given by: $m = \frac{y_2 - y_1}{x_2 - x_1}$

   For points P(-2, -1) and Q(3, 1):

   $m_{PQ} = \frac{1 - (-1)}{3 - (-2)} = \frac{2}{5}$

2. Slope of QR: For points Q(3, 1) and R(1, 6):

   $m_{QR} = \frac{6 - 1}{1 - 3} = \frac{5}{-2} = -\frac{5}{2}$

3. Relation between Slopes: The product of slopes of two perpendicular lines is -1.

   $m_{PQ} * m_{QR} = \frac{2}{5} * -\frac{5}{2} = -1$

Since the product of slopes is -1, QR is perpendicular to PQ, hence QR is tangent to the circle c.