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The circle $C$ has equation $$x^2-8x+y^2+6y=24$$ I completed the square: $$(x-4)^2+(y+3)^2=49$$ Therefore the radius is 7. I want to know if the point $(9,2)$ is inside the circle. Substituting into the distance formula: $$\sqrt{(9-4)^2 + (2+3)^2}=\sqrt{50}=7.07\dots$$ So the point lies outside the circle, or have I gone wrong?

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    $\begingroup$ $50>49$ outside, yes! $\endgroup$ – Nosrati Jan 10 '17 at 10:08
  • $\begingroup$ Thank you :) I always worry my answer is wrong $\endgroup$ – user405455 Jan 10 '17 at 10:09
  • $\begingroup$ I think your workings are fine. $\endgroup$ – Rohan Jan 10 '17 at 10:09
  • $\begingroup$ you are correct $\endgroup$ – Bhaskara-III Jan 10 '17 at 11:09
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    $\begingroup$ While many are saying that your answer is correct but over-complicated, I would argue that your answer is probably exactly what your teacher is looking for. $\endgroup$ – Gorchestopher H Jan 10 '17 at 15:34
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Perfectly fine.

However, do note you didn't actually need to compute the square root. Since for all positive real numbers $x>y$ if and only if $x^2>y^2$, it's enough to observe that $50>49$, and that therefore $\sqrt{50}>\sqrt{49}$.

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Too complicated. You don't need to complete the square of $$x^2-8x+y^2+6y=24$$ Bring it in a form such that $x^2$ and $y^2$ are on the same side of the equation and that they both have both a positive coefficient. If this is not possible, it is not the equation of a circle.

Your equation is already in such a form.

Now plugin the coordinates of your point. If equality holds for your equation then the point lies on the circle. If the side that contained $x^2$ and $y^2$ is the larger then the point lies in the outside of the circle. If the side that contained $x^2$ and $y^2$ is the smaller one then the point lies in the inner of the circle.

Example:

Point$(9,2)$: $$9^2-8\cdot 9 + 2^2 +6 \cdot 2 = 25$$ and $25$ is greater than $24$, so the point lies on the outside.

You can bring it to a form where the right side is $0$ and the coefficients of $x^2$ and $y^2$ (on the left side) is $1$. This representation is unique (except the order of the terms because of commutativity of the $+$ operator). And now the rule is simple: If you plug in the point and the LHS (left hand side) is $0$ then the point lies on the circle. If the LHS is negative then the point lies in the inner of the circle. If the LHS is positive then the point lies on the outside of the circle.

Example:

The LHS of the circle equation $$x^2-8x+y^2+6y-24=0$$ is 1 if you plug in $(9,2)$. So the point lies in the outside.

The left hand side is the expansion of the expression

$$((x-c_x)^2+(y-c_y)^2)-r^2$$

The first term is the sqare of the distance of the Point $(x,y)$ to the center $(c_x,c_y)$ of the circle, the second is the square of the radius. From this immediately follow the preceding statements.

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    $\begingroup$ -1 Your method requires an argument which you don't give. Why does this work? Why is the point outside the circle if this expression is positive, and inside otherwise? OP's answer is simple and very clear from the geometric picture. Why would you call that "too complicated"? $\endgroup$ – Servaes Jan 10 '17 at 19:05
  • $\begingroup$ @Servaes It is more complicated because completing the square and plug in the values (and calculating a root) needs more effort and is more error prone than only plugging in the values. I assume that this will not be the only circle that the OP will meet during his mathematical sudies, So it makes sebse to point him to a more efficient technique. I added a proof that shows why this works at the end. $\endgroup$ – miracle173 Jan 10 '17 at 22:54
  • $\begingroup$ It seems we disagree on the meaning of complicated. Thanks for adding a proof though. Note that your proof is precisely the approach OP took. Also, other answer already pointed out that there is no need to compute a square root. $\endgroup$ – Servaes Jan 10 '17 at 22:55
  • $\begingroup$ @Servaes It seems so. Note that I put the statement concerning the square root in paranthesis. I did this because it isn't necessary to calculate it. $\endgroup$ – miracle173 Jan 10 '17 at 23:10
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You are correct, as $$7.07\gt 7$$ and since the radius of the circle is $7$ the point $(9,2)$ must lie (just) outside the circle.

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You are correct, since the distance of point from the center is more than the that of radius i.e. $$7<7.07$$

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Too complicated.

Simply re-write the given equation in the form $f(x, y) = 0$

That is, $x^2-8x+y^2+6y-24 = 0$

$d = \sqrt {f(x_o, y_o)}$ is the "length" of tangent from $P(x_o, y_o)$ to $C:f(x, y) = 0$.

Revised:-

If ${f(x_o, y_o)} \gt 0$, that tangent is a real one and P is outside of $C$.

If ${f(x_o, y_o)} \lt 0$, that tangent is an imaginary one and P is inside $C$.

If ${f(x_o, y_o)} = 0$; guess what.......

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  • $\begingroup$ You mean $f(x_0, y_0) < 0$. $d$ is either $\ge 0$ or is imaginary. $\endgroup$ – Paul Sinclair Jan 10 '17 at 17:17
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    $\begingroup$ -1 For calling OP's clear geometric argument that is accesible to any 12-year old "Too complicated", and proceeding with a rather vague and completely unexplained and unmotivated trick using real and imaginary tangents to implicit curves. $\endgroup$ – Servaes Jan 10 '17 at 19:11
  • $\begingroup$ @PaulSinclair Revised. Thanks. $\endgroup$ – Mick Jan 11 '17 at 17:13
  • $\begingroup$ @Servaes Whether it is a trick or an advanced technique depends on how one looks at it. Time required to solve the problem may be a tool for such measurement. $\endgroup$ – Mick Jan 11 '17 at 17:15
  • $\begingroup$ @Mick IMHO the difference between a trick and an advanced technique is not in the method but in the user. It is the difference between imitating and understanding a method. This answer does not show or provide any understanding of the problem or the method used to solve it, which is why I find it a poor answer. But that's just my take on it. $\endgroup$ – Servaes Jan 11 '17 at 17:46

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