Answer:
[tex] \begin{cases} (x - 2)^2 + (y - 2)^2 = 25 \\ y = 5 \end{cases} [/tex]
Solutions: x = 6, y = 5 or x = -2, y = 5
Step-by-step explanation:
Use a graph.
Plot point (-2, 5). That will be a point on a circle with radius 5.
From point (-2, 5), go right 4 and down 3 to point (2, 2). (2, 2) is the center of the circle.
You now need the equation of a circle with center (2, 2) and radius 5.
Use the standard equation of a circle:
[tex] (x - h)^2 + (y - k)^2 = r^2 [/tex]
where (h, k) is the center and 5 is the radius.
The circle has equation:
[tex] (x - 2)^2 + (y - 2)^2 = 25 [/tex]
To have a single solution, you need the equation of the line tangent to the circle at (-2, 5), but since you want more than one solution, you need the equation of a secant to the circle. For example, use the equation of the horizontal line through point (2, 5) which is y = 5.
System:
[tex] \begin{cases} (x - 2)^2 + (y - 2)^2 = 25 \\ y = 5 \end{cases} [/tex]
To solve, let y = 5 in the equation of the circle.
(x - 2)^2 + (5 - 2)^2 = 25
(x - 2)^2 + 9 = 25
(x - 2)^2 = 16
x - 2 = 4 or x - 2 = -4
x = 6 or x = -2
Solutions: x = 6, y = 5 or x = -2, y = 5
