MysteryMoon
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Prove that 3x + 4y +21 = 0 is a tangent to the circle x^2 -8x +y^2 + 4y - 5 = 0
Another help with locus question? (1 Viewer)
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leehuan
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Firstly, since you eventually had 3 questions, you probably could've just posted them in one thread. I think the forum members are generally nice but it can be viewed as spamming.
Anyway, as for the question:
I'm not entirely sure if this is locus, because there is actually a trick out of the quadratics topic that can help you here. The idea is:
Rearranging 3x+4y=21 -> y=-21/4 - 3x/4 (or x=-7-4y/3 works too)
You can sub this into the equation of the circle
(Note that there will be messy algebra as a result of all of this)
The resulting expression can be used to find any points of intersection.
And here's where you need to really think:
If you simplify the expression out, you'll end up with a quadratic. This quadratic can either have no solutions, have one solution, or have two solutions.
But the idea is:
If the quadratic has no solutions, then the line never touches the circle - this is because there are NO points of intersection
If the quadratic has two solutions, then the line cuts the circle twice; it is a secant to the circle - this is because there are TWO points of intersection
However if the quadratic only has one solution, then the line must be tangent to the circle. Note that for a circle, any line that is tangent will never cut the circle again.
Therefore, by proving that the discriminant of this quadratic equals 0, you show that it must be tangent. This is due to the fact that if the quadratic discriminant equals 0, we know there is only one (real) solution.
 }^{ 2 }+4\left( -\frac { 3x }{ 4 } -\frac { 21 }{ 4 } \right) -5=0)
Anyway, as for the question:
I'm not entirely sure if this is locus, because there is actually a trick out of the quadratics topic that can help you here. The idea is:
Rearranging 3x+4y=21 -> y=-21/4 - 3x/4 (or x=-7-4y/3 works too)
You can sub this into the equation of the circle
(Note that there will be messy algebra as a result of all of this)
The resulting expression can be used to find any points of intersection.
And here's where you need to really think:
If you simplify the expression out, you'll end up with a quadratic. This quadratic can either have no solutions, have one solution, or have two solutions.
But the idea is:
If the quadratic has no solutions, then the line never touches the circle - this is because there are NO points of intersection
If the quadratic has two solutions, then the line cuts the circle twice; it is a secant to the circle - this is because there are TWO points of intersection
However if the quadratic only has one solution, then the line must be tangent to the circle. Note that for a circle, any line that is tangent will never cut the circle again.
Therefore, by proving that the discriminant of this quadratic equals 0, you show that it must be tangent. This is due to the fact that if the quadratic discriminant equals 0, we know there is only one (real) solution.
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