Matha problem: how to do (b) i and ii, thanks?

(a)

L passes through (15, 5) and (12, 0), so its slope is (5 - 0)/(15 - 12) = 5/3.

L' is perpendicular to L, so its slope is -3/5.

L' also passes through Q(12, 0) because it is the point of intersection with L on the x-axis.

The equation of L' is

(y - 0)/(x - 12) = -3/5

5y = 36 - 3x

3x + 5y - 36 = 0

(b)

(i)

The y-coordinate of G is given as k and G is on L'.

Therefore, we can find the x-coordinate of G by putting y = k in L'.

3x + 5k - 36 = 0

3x = 36 - 5k

x = (36 - 5k)/3

That is, G = ( (36 - 5k)/3, k ), which is the centre of circle C.

Circle C passes through Q(12, 0), so QG is the radius.

Therefore, the equation of C is

[x - (36 - 5k)/3]² + (y - k)² = QG² = [12 - (36 - 5k)/3]² + (0 - k)²

x² - 2x(36 - 5k)/3 + (36 - 5k)²/3² + y² - 2yk + k² = 144 - 24(36 - 5k)/3 + (36 - 5k)²/3² + k²

x² - 2x(36 - 5k)/3 + y² - 2yk = 144 - 24(36 - 5k)/3

3x² - 2x(36 - 5k) + 3y² - 6yk = 432 - 24(36 - 5k)

3x² - 2x(36 - 5k) + 3y² - 6yk = 432 - 864 + 120k

3x² + 3y² - 2(36 - 5k)x - 6ky + 432 - 120k = 0

(ii)

C passes through A(4, 8), so put x = 4 and y = 8 into C to solve for k:

3(4²) + 3(8²) - 2(36 - 5k)(4) - 6k(8) + 432 - 120k = 0

k = 3

Therefore, G = (7, 3).

Note that P = (15, 5).

The smallest circle which passes through P and G would be the one with PG being its diameter.

This diameter is √[ (15 - 7)² + (5 - 3)² ] = √(64 + 4) = √68 = 2√17.

The radius of this smallest circle is √17.

The area of this smallest circle is π × (√17)² = 17π.