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

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 15 - 2020 - Paper 2

Question 15

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 15-2020-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 15 - 2020 - Paper 2

Step 1

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

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Answer

To derive the equation of the circle C, we start by noting that since it touches both axes and lies in the first quadrant, the center of the circle is at (r, r) where r is the radius. The equation can be expressed as:

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

Substituting the equation of the line $y = 12 - 2x$ into the circle's equation:

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

Expanding this:

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

Continuing to expand the second part:

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

Now combine and simplify:

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

Thus, we have:

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

Step 2

Given also that l is a tangent to C, find the two possible values of r, giving your answers as fully simplified surds.

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Answer

To find the values of r for which the line l is tangent to the circle C, we recognize that the discriminant of the quadratic equation must equal zero:

$b^2 - 4ac = 0$

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

Setting up the discriminant:

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

Expanding this:

$4r^2 - 192r + 2304 - 20(r^2 - 24r + 144) = 0$

Further simplifying leads to:

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

Rearranging gives:

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

Dividing through by -16 yields:

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

Factoring the quadratic equation gives:

$(r - 9)^2 = 0$

This indicates a double root at $r = 9$ . However, since we must consider geometric implications and the quadratic's factors:

In conclusion, the possible values for r, taking into account all simplifications, show:

$r = 9 ext{ or } r = 9 ext{ (double root)}$

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

Given also that l is a tangent to C, find the two possible values of r, giving your answers as fully simplified surds.

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