The circle c has centre P(-2, -1) and passes through the point Q(3, 1) - Leaving Cert Mathematics - Question 4 - 2013

To find the radius of circle c:

1. Calculate the distance from center P(-2, -1) to point Q(3, 1) using the distance formula:

   $r = ext{distance}(P, Q) = \\sqrt{(x_Q - x_P)^2 + (y_Q - y_P)^2} = \\sqrt{(3 - (-2))^2 + (1 - (-1))^2} = \\sqrt{(5)^2 + (2)^2} = \\sqrt{25 + 4} = \\sqrt{29}$

2. The radius r is thus ( r = \sqrt{29} ).

3. The equation of the circle is given by

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

To determine if QR is a tangent to circle c:

1. Calculate the slope of line segment PQ:

   • Using the slope formula ( m = \frac{y_2 - y_1}{x_2 - x_1} ):

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

2. Determine the coordinates of point R, which is (1, 6).

3. Calculate the slope of segment QR:

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

4. For a tangent line, the product of the slopes of the two lines (PQ and QR) should equal -1:

   $m_{PQ} \times m_{QR} = \frac{2}{5} \times \left(-\frac{5}{2}\right) = -1$

5. This calculation confirms that line QR is tangential to circle c.