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

The original circle C has its centre at the origin (0, 0) with a radius of 4 (since (x^2 + y^2 = 16) implies a radius of (r = 4)). After translating by the vector (egin{pmatrix} 3 \ 0 \ 0 \end{pmatrix}), the new centre for circle B becomes (3, 0).

• Centre of Circle B: (3, 0)

To find the points of intersection of circle B with the x-axis, we set (y = 0) in the equation of circle B:

The equation of circle B, after the translation, is:

$(x - 3)^2 + y^2 = 16$

Setting (y = 0):

$(x - 3)^2 = 16$

Taking square roots gives:

$x - 3 = 4 \quad \text{or} \quad x - 3 = -4$

Thus,

• $x = 7 \implies (7, 0)$
• $x = -1 \implies (-1, 0)$

The points of intersection with the x-axis are (7, 0) and (-1, 0).