Question

solve the system:
\

$$\begin{cases} (x - 1)^2 + y^2 = 25 \\\\ x - y^2 = -4 \\end{cases}$$

(-4, ? ) and ( , -3) and ( , )

Explanation:

Step1: Solve for \( x \) from the second equation

From \( x - y^2 = -4 \), we can express \( x \) as \( x = y^2 - 4 \).

Step2: Substitute \( x = y^2 - 4 \) into the first equation

Substitute \( x \) into \( (x - 1)^2 + y^2 = 25 \):

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So \( y^2 = 0 \) or \( y - 3 = 0 \) or \( y + 3 = 0 \), which gives \( y = 0 \), \( y = 3 \), or \( y = -3 \).

Step3: Find \( x \) for each \( y \)

• When \( y = 0 \): \( x = 0^2 - 4 = -4 \), so the point is \( (-4, 0) \).
• When \( y = 3 \): \( x = 3^2 - 4 = 5 \), so the point is \( (5, 3) \).
• When \( y = -3 \): \( x = (-3)^2 - 4 = 5 \), so the point is \( (5, -3) \).

Answer:

For the point \((-4, [?])\), the \( y \)-value is \( 0 \). For the point \(([?], -3)\), the \( x \)-value is \( 5 \). The third point is \((5, 3)\). So the solutions are \((-4, 0)\), \((5, -3)\), and \((5, 3)\).