The circle c has centre P(−2, −1) and passes through the point Q(3, 1) - Leaving Cert Mathematics - Question 4 - 2014

To find the radius, we calculate the distance from center P(-2, -1) to point Q(3, 1) using the distance formula:

$r = ext{distance}(P, Q) = \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Substituting the points:

$r = \\sqrt{(3 - (-2))^2 + (1 - (-1))^2} = \\sqrt{(3 + 2)^2 + (1 + 1)^2} = \\sqrt{(5)^2 + (2)^2} = \\sqrt{25 + 4} = \\sqrt{29}$

Thus, the radius of the circle c is $\sqrt{29}$, and the equation of the circle is:

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

To show that QR is a tangent to the circle, we first find the slopes of lines PQ and QR.

1. Calculating the slope of PQ: $m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-1)}{3 - (-2)} = \frac{2}{5}$

2. Now calculate the slope of QR: $m_{QR} = \frac{6 - 1}{1 - 3} = \frac{5}{-2} = -\frac{5}{2}$

3. Since QR is tangent to the circle at point Q, the slopes should satisfy: $m_{PQ} \times m_{QR} = -1$ Substitute the slopes: $\left(\frac{2}{5}\right) \times \left(-\frac{5}{2}\right) = -1$ Thus, the slopes are negative reciprocals, confirming QR is tangent to the circle c.