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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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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 $\sqrt{50}>\sqrt{49}$.

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.

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.

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

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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