A-Level Edexcel Maths Pure Quadratics: 14. A circle C with radius r

14. 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 14 - 2020 - Paper 2

Question 14

14.-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 14-2020-Paper 2.png

14. 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 coo... show full transcript

Worked Solution & Example Answer:14. 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 14 - 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 find the points of intersection, we first express y in terms of x using the line equation:

$y = 12 - 2x$

Now, since the circle C touches both the x-axis and y-axis, its center must be at the point (r, r). Using the standard form of a circle, we have:

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

Substituting for y, we get:

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

Expanding this, we have:

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

This simplifies to:

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

Continuing with the simplification gives:

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

This can be rearranged to yield:

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

Step 2

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

99%

104 rated

Answer

Given that the line l is a tangent to the circle C, we can set the discriminant of the quadratic to zero:

$b^2 - 4ac = 0$

Using coefficients from our earlier derived equation, we assign:

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

The discriminant becomes:

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

Expanding this equation:

$(2r - 48)^2 = 100(r^2 - 24r + 144)$

Solving leads to:

$4r^2 - 192r + 2304 = 100r^2 - 2400r + 14400$

Rearranging terms:

$96r^2 - 720r + 12096 = 0$

Simplifying:

$r^2 - 7.5r + 126 = 0$

Using the quadratic formula:

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

Substituting the values gives:

$r = \frac{7.5 \pm \sqrt{(7.5)^2 - 4(1)(126)}}{2(1)}$

This results in:

$r = \frac{7.5 \pm 9\sqrt{3}}{2}$

Thus, the two possible values for r are:

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

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...