A-Level Edexcel Maths Pure Proof: A circle C with radius r

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 16 - 2020 - Paper 2

Question 16

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 16-2020-Paper 2.png

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 coordi... show full transcript

Worked Solution & Example Answer: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 16 - 2020 - Paper 2

Step 1

Show that the x coordinates of the points of intersection of l with C satisfy

96%

114 rated

Answer

To derive the equation of the circle C, we first note that it touches the x-axis and y-axis, meaning the center is at (r, r). The general form of the circle equation centered at (h, k) with radius r is:

$(x - r)^2 + (y - r)^2 = r^2$

Substituting (h, k) with (r, r):

$(x - r)^2 + (y - r)^2 = r^2$

Next, we can express y in terms of x using the equation of the line:

$y = 12 - 2x$

Substituting this in the circle equation gives:

$(x - r)^2 + (12 - 2x - r)^2 = r^2$

Expanding this results in:

$x^2 - 2xr + r^2 + (12 - 2x - r)^2 = r^2$

Next, simplify the right side,

$(12 - 2x - r)^2 = 144 - 48x + 4x^2 - 24r + 24x + r^2$

Combining all terms,

$x^2 - 2xr + r^2 + 4x^2 - 48x + 144 - 24r = 0$

Collect like terms to derive:

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

Step 2

find the two possible values of r, giving your answers as fully simplified surds.

99%

104 rated

Answer

Since the line l is tangent to circle C, the discriminant of the quadratic must equal 0. The equation is:

$b^2 - 4ac = 0$

Where in our equation:

a = 5,
b = (2r - 48),
c = (r^2 - 24r + 144).

Substituting these gives:

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

Expanding this leads to:

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

Combining terms results in:

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

Dividing all terms by -16:

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

Solving this quadratic using the quadratic formula yields:

$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{18 \pm \sqrt{(18)^2 - 4(1)(36)}}{2(1)}$ $= \frac{18 \pm \sqrt{324 - 144}}{2}$ $= \frac{18 \pm \sqrt{180}}{2}$ $= \frac{18 \pm 6\sqrt{5}}{2}$ $= 9 \pm 3\sqrt{5}$

Thus the two possible values for r are:

$r = 9 + 3\sqrt{5}$ and $r = 9 - 3\sqrt{5}$ .

Show that the x coordinates of the points of intersection of l with C satisfy

find the two possible values of r, giving your answers as fully simplified surds.

Join the A-Level students using SimpleStudy...