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August 28th, 2013, 05:33 AM   #1
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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 a tangent to the circle ?

Attempt: Centre c = (-2, 2) & radius
If T is a tangent, the distance between c & T

or 3
Solving yields imaginary numbers.
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August 28th, 2013, 07:08 AM   #2
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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.
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August 28th, 2013, 07:18 AM   #3
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Re: Coordinate Geometry - The Circle

Ok, thank you.
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