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 August 28th, 2013, 05:33 AM #1 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Coordinate Geometry - The Circle My answer to this question is in line with that of the text book, but I just wanted to make sure if I was correct about applying the square to the line marked with . Specifically, does squaring both sides of the problem allow me to remove the modulus and, as such, should I remove the modulus from the expression marked with , rather than doing so in the following lines? Many thanks. Q. For what values of k is the line $2x-ky-3=0$ a tangent to the circle $x^2+y^2+4x-4y-5=0$? Attempt: Centre c = (-2, 2) & radius $=\sqrt{(-2)^2+2^2+5}=\sqrt{13}$ If T is a tangent, the distance between c & T $=\frac{|-2(2)+2(-k)-3|}{\sqrt{2^2+(-k)^2}}=\sqrt{13}$ $\frac{|-4-2k-3|}{\sqrt{4+k^2}}=\sqrt{13}$ $\frac{|-7-2k|}{\sqrt{4+k^2}}=\sqrt{13}$ $\frac{|-7-2k|^2}{4+k^2}=13$ $|4k^2+28k+49|=13k^2+52$ $4k^2+28k+49=+/-13k^2+52$ $4k^2+28k+49=13k^2+52$ $9k^2-28k+3\,=0$ $k=\frac{1}{9}$ or 3 Solving $4k^2+28k+49=-13k^2-52$ yields imaginary numbers.
 August 28th, 2013, 07:08 AM #2 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Coordinate Geometry - The Circle Re the highlighted step. As IxI^2 = x^2, you could have got rid of the modulus at that point and avoided worrying about the -ve possibility. Another way to do it is to look for values of k for which the line meets to circle at precisely one point. The line will either miss the circle, meet it at two points (a chord) or at one point (if the line is a tangent). We have: x = 0.5(ky + 3), which can be plugged into the equation for the circle, giving: (k^2 +4)y^2 + (14k - 16)y + 13 = 0 This will have precisely one solution for y if: (14k - 16)^2 - 4(k^2 + 4)13 = 0, which yields a quadratic in k: 9k^2 - 28k + 3 = 0 Hence, k = 3 or 1/9.
 August 28th, 2013, 07:18 AM #3 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Re: Coordinate Geometry - The Circle Ok, thank you.

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