The circle c has equation $(x + 2)^{2} + (y - 3)^{2} = 100.$ Write down the co-ordinates of A, the centre of c - Leaving Cert Mathematics - Question (a) - 2014

The radius r can be deduced from the equation of the circle. The right-hand side of the equation gives us the radius squared. Thus, we have:

$r = ext{sqrt}(100) = 10.$

To verify that the point P(-8, 11) lies on the circle, we substitute the coordinates into the circle equation:

$(-8 + 2)^{2} + (11 - 3)^{2} = 100.$

Calculating each term gives:

$(-6)^{2} + (8)^{2} = 36 + 64 = 100.$

Since both sides equal 100, this confirms that the point P is indeed on the circle.

To find the slope of the radius [AP], we use the coordinates of points A and P:

ext{slope} = rac{y_{2} - y_{1}}{x_{2} - x_{1}} = rac{11 - 3}{-8 + 2} = rac{8}{-6} = -rac{4}{3}.

The slope of the tangent line t will be the negative reciprocal of the radius slope:

ext{slope}_{t} = rac{3}{4}.

Using point-slope form with point P(-8, 11), the equation of the tangent line can be expressed as:

y - 11 = rac{3}{4}(x + 8).

Rearranging to slope-intercept form gives: y = rac{3}{4}x + 8.

Since line k is parallel to the tangent and goes through point Q, we know it has the same slope. Using point-slope form again with point Q:

y - 11 = rac{3}{4}(x + 8).

We find the coordinates of Q when setting $x = 4, ext{ which gives } Q(4, -5).$ Thus, the co-ordinates of Q are $Q(4, -5).$