The circle c has equation $(x - 1)^2 + (y + 4)^2 = 25$ - Leaving Cert Mathematics - Question 4 - 2020

Given the point (1, k) is on the circle:

$(1 - 1)^2 + (k + 4)^2 = 25$

This simplifies to:

the equation becomes $(k + 4)^2 = 25$

Taking the square root of both sides, we get:

$k + 4 = 5 \\ k + 4 = -5$

Solving these gives:

• From $k + 4 = 5$: $k = 1$
• From $k + 4 = -5$: $k = -9$

Thus, the two possible values of k are:

• k = 1
• k = -9

The equation of the circle is given as:

$x^2 + y^2 = 13$

The coordinates of point A are (3, -2). To find the slope of the radius at A:

$m_{radius} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 0}{3 - 0} = \frac{-2}{3}$

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

$m_{tangent} = \frac{3}{2}$

Using the point-slope formula for the tangent line:

$y - y_1 = m(x - x_1)$

Substituting the coordinates of A and the slope:

$y - (-2) = \frac{3}{2}(x - 3)$

Simplifying this:

$y + 2 = \frac{3}{2}x - \frac{9}{2}$

Rearranging gives:

$3x - 2y - 13 = 0$

Thus, the equation of the tangent t in the required form is:

$3x - 2y + 13 = 0$