A circle C with radius r - lies only in the 1st quadrant - touches the x-axis and touches the y-axis The line l has equation 2x + y = 12 (a) Show that the x coordinates of the points of intersection of l with C satisfy 5x² + (2r - 48)x + (r² - 24r + 144) = 0 (b) Given also that l is a tangent to C, find the two possible values of r, giving your answers as fully simplified surds. - Edexcel - A-Level Maths Pure - Question 1 - 2019 - Paper 2

Since line l is tangent to the circle, we set the discriminant of the quadratic equation to zero:

$b^2 - 4ac = 0$

where:

• $a = 5$
• $b = (2r - 48)$
• $c = (r^2 - 24r + 144)$

Calculating the discriminant gives:

$(2r - 48)^2 - 4(5)(r^2 - 24r + 144) = 0$

This simplifies to:

$(4r^2 - 192r + 2304 - 20r^2 + 480r - 2880) = 0$

Leading to:

$-16r^2 + 288r - 576 = 0$

Factoring out -16 gives:

$r^2 - 18r + 36 = 0$

Using the quadratic formula:

$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

yields: $r = \frac{18 \pm \sqrt{(18)^2 - 4 \cdot 1 \cdot 36}}{2}$

which simplifies to: $r = \frac{18 \pm \sqrt{324 - 144}}{2}$ This gives: $r = \frac{18 \pm \sqrt{180}}{2} = \frac{18 \pm 6\sqrt{5}}{2}$ Thus, the possible values of r are:

$r = 9 \pm 3\sqrt{5}$