Question

14 a circle with center c(-3, 3) is graphed on a coordinate plane. the circle is transformed so that its image forms a circle with center c(-3, -3). which rule does not describe a transformation that results in a circle with center c? a rotation 90° counterclockwise: (x, y) → (-y, x) b translation 6 units down: (x, y) → (x, y - 6) c reflection across the x-axis: (x, y) → (x, -y) d reflection across the y-axis: (x, y) → (-x, y)

Explanation:

Step1: Analyze Option A (Rotation)

Apply the rotation rule \((x, y) \to (-y, x)\) to \(C(-3, 3)\).
Substitute \(x = -3\), \(y = 3\): \((-3, 3) \to (-3, -3)\)? Wait, no: \(-y = -3\), \(x = -3\)? Wait, no: \(x=-3\), \(y = 3\), so \(-y=-3\), \(x=-3\)? Wait, no, the rule is \((x,y)\to (-y, x)\), so \(x=-3\), \(y = 3\) becomes \((-3, -3)\)? Wait, no: \(-y = -3\), \(x=-3\)? Wait, no, \(x=-3\), \(y = 3\), so \(-y=-3\), \(x=-3\)? Wait, no, the rule is \((x,y)\to (-y, x)\), so \(x=-3\), \(y = 3\) gives \((-y, x)=(-3, -3)\)? Wait, no, \(y = 3\), so \(-y=-3\), \(x=-3\), so \((-3, -3)\). Wait, that's \(C'\). Wait, maybe I miscalculated. Wait, \(x=-3\), \(y = 3\): \(-y = -3\), \(x=-3\), so \((-3, -3)\). So rotation gives \(C'\).

Step2: Analyze Option B (Translation)

Apply translation \((x, y) \to (x, y - 6)\) to \(C(-3, 3)\).
Substitute: \(x=-3\), \(y = 3\), so \(y - 6 = 3 - 6 = -3\). So \((-3, -3)\), which is \(C'\).

Step3: Analyze Option C (Reflection over x - axis)

Apply reflection rule \((x, y) \to (x, -y)\) to \(C(-3, 3)\).
Substitute: \(x=-3\), \(y = 3\), so \(-y=-3\). So \((-3, -3)\), which is \(C'\).

Step4: Analyze Option D (Reflection over y - axis)

Apply reflection rule \((x, y) \to (-x, y)\) to \(C(-3, 3)\).
Substitute: \(x=-3\), so \(-x = 3\), \(y = 3\). So \((3, 3)\), which is not \(C'(-3, -3)\).

Answer:

D. Reflection across the \(y\)-axis: \((x, y) \to (-x, y)\)