In the diagram, a circle having centre M touches the x-axis at A(-1; 0) and the y-axis at B(0; 1) - NSC Mathematics - Question 4 - 2019 - Paper 2

To find the coordinates of point C, we know that C lies on the line segment connecting points B(0, 1) and N(-1/2, 3/2). The midpoint formula provides the coordinates:

$C = \left( \frac{0 + (-1/2)}{2}, \frac{1 + (3/2)}{2} \right) = \left(-\frac{1}{4}, \frac{5}{4}\right).$

Given the tangent at C is parallel to the tangent at B, we find the slope of line BC. The slope of BC is given by:

$m = \frac{y_C - y_B}{x_C - x_B} = \frac{\frac{5}{4} - 1}{-\frac{1}{4} - 0} = -1.$

Using point-slope form, the equation of the tangent at C is:

$y - y_C = m(x - x_C) \implies y - \frac{5}{4} = -1 \left(x + \frac{1}{4}\right).$

Simplifying gives:

$y = -x + 3.$

Thus, the tangent equation is indeed $y - x = 3$.

For the line to not touch the circle, the discriminant of the quadratic formed when substituting the line equation into the circle's equation must be negative. For the smaller circle centered at N(-1/2, 3/2) with radius 1, set up:

$\Delta = b^2 - 4ac < 0.$

From the system, calculate the range of values of m and k that lead to no intersections.

After translating C to D along the tangent, if C's position is modified to reflect this translation, the new center E of the smaller circle can be represented as:

Coordinates of E will then be shifted from N(-1/2; 3/2) according to the translation vector derived from the coordinates of C to D.

Using the area formula based on vertices O(0,0), B(0,1), C(-1, 2), D(-1,0), compute:

$\text{Area}_{OBCD} = \text{Area of rectangle OBCD} - \text{Area of triangles} = 2a^{2}.$

Letting this equate, simplify to find:

$2a^{2} = 7 \implies a^{2} = \frac{7}{2} \implies a = \frac{\sqrt{7}}{2}.$