The circle c has equation $x^{2} + y^{2} - 4x + 2y - 4 = 0.$ The point A is the centre of the circle - Leaving Cert Mathematics - Question 2 - 2020

Using the distance formula to find the distance from point A(2, -1) to point B(5, 8):

$d = ext{sqrt}((x_2 - x_1)^{2} + (y_2 - y_1)^{2})$. Substituting the coordinates: $d = ext{sqrt}((5 - 2)^{2} + (8 - (-1))^{2})$ $d = ext{sqrt}(3^{2} + 9^{2})$ $d = ext{sqrt}(9 + 81)$ $d = ext{sqrt}(90) = 3 ext{sqrt}(10)$. Thus, $|BT| = 3 ext{sqrt}(10)$.

Both circles c₁ and c₂ have their centres on the x-axis, and their radius is 5 units. Since the point (1, 4) lies on both circles, let's denote the centres of c₁ and c₂ as (g, 0) and (h, 0), respectively.

1. From the distance formula: $(1 - g)^{2} + (4 - 0)^{2} = 25$. This simplifies to: $(1 - g)^{2} + 16 = 25$, which gives: $(1 - g)^{2} = 9$, leading to: $1 - g = 3 ext{ or } 1 - g = -3$.

Thus:

• For $g = -2$ (Circle c₁)
• For $g = 4$ (Circle c₂)

2. Now, we write the equations:
   • Circle c₁: $(x + 2)^{2} + y^{2} = 25$
   • Circle c₂: $(x - 4)^{2} + y^{2} = 25$.