The equation of a curve is $y = ax^2$ - Edexcel - GCSE Maths - Question 20 - 2017 - Paper 3

Circle C has a center at the origin $(0, 0)$ and a radius of 4 (since $\sqrt{16} = 4$). After translating circle C by the vector (\begin{pmatrix} 3 \ 0 \end{pmatrix}), the center of circle B moves to $(3, 0)$.

To sketch circle B:

• Draw a circle with center $(3, 0)$ and radius 4.
• The points of intersection with the x-axis can be found by setting $y = 0$ in the equation of the circle.

The equation of circle B is: $(x - 3)^2 + y^2 = 16$ Setting $y = 0$ gives: $(x - 3)^2 = 16$
Taking the square root leads to: $x - 3 = \\pm4$ Thus, the intersection occurs at $x = 7$ and $x = -1$.

The labeled points are:

• Center: $(3, 0)$
• Points of intersection: $(7, 0)$ and $(-1, 0)$.