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The epsilon-delta definition of a limit is an algebraically precise formulation of evaluating the limit of a function. Informally, the definition states that a limit $L$ of a function at a point $x_0$ exists if no matter how $x_0$ is approached, the values returned by the function will always approach $L$. This definition is consistent with methods used to evaluate limits in elementary calculus, but the mathematically rigorous language associated with it appears in higher level analysis. The $\varepsilon$-$\delta$ definition is also useful when trying to show the continuity of a function.

The equation of any conic can be expressed as

$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0.$

However, the condition for the equation to represent a circle is $a = b$ and $h = 0$. Then the general equation of the circle becomes

$x^2 + y^2 + 2gx + 2fy + c = 0.$

Unfortunately, it can be difficult to decipher any meaningful properties about a given circle from its general equation, so completing the square allows quick conversion to the standard form, which contains values for the center and radius of the circle.

Convert $x^2 + y^2 - 4x + 6y - 23 = 0$ into standard form by completing the square.

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First, we group the $x$ terms together and the $y$ terms together while adding 23 to both sides, giving us

$x^2 - 4x + y^2 + 6y = 23.$

When we complete the square for both groups of $x$ and $y$ terms, we obtain the standard form

$(x - 2)^2 + (y + 3)^2 = 23 + 4 + 9 = 36. \ _\square$

This is the standard equation of a circle, with radius $r$ and center at $(a,b)$.

The general form can be converted into the standard form by completing the square. First, we combine like terms

$\begin{aligned} x^2 + 2gx + y^2 + 2fy + c &= 0\\ (x+g)^2 + (y+f)^2 - g^2 - f^2 + c &= 0\\ (x+g)^2 + (y+f)^2 &= g^2 + f^2 - c. \end{aligned}$

Comparing this with the standard form, we come to know that

• center $= (-g, -f)$
• radius $= \sqrt{g^2 + f^2 -c}.$

However, make sure the coefficients of $x^2$ and $y^2$ are 1 before applying these formulae.

While solving problem we try to make left hand side of the form $(x-a)^2+(y-b)^2$ by using completing the square method.

Draw the graph of the equation $(x-3)^2+(y-5)^2=49$.

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Notice that the right side is $7^2$. Comparing to the standard equation of circle we easily see that the graph is a circle with radius $7$ and center at $(3,5)$. Now we can easily draw the graph using compass. $_\square$

What does the graph of the equation $x^2+y^2-2x-14y+34=0$ look like?

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Notice that we can rewrite the equation as

$\left(x^2-2x+1\right)+\left(y^2-14y+49\right)=16.$

Completing the squares, this becomes

$(x-1)^2+(y-7)^2=4^2.$

So the graph is a circle with radius $4$, centered at $(1,7)$. $_\square$

What are the radius and center of the circle whose equation is $x^2-4x+y^2+2x=44$?

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We can apply completing the square method to left hand side:

$\begin{aligned} x^2-4x+2^2+y^2+2x+1^2&=49+2^2+1^2\\ (x^2-4x+4)+(y^2+2x+1)&=49\\ (x-2)^2+(y+1)^2&=7^2. \end{aligned}$

Comparing with the standard equation, we can see that $a=2$ and $b=-1.$ Therefore, the center of the circle is $(2,-1)$ and its radius is $7.$ $_\square$

What are the radius and center of the circle whose equation is $x^2+y^2=25$?

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We can rewrite the given equation as

$(x-0)^2+(y-0)^2=5^2.$

Comparing with the standard equation, we can see that $a=b=0.$ Therefore the center of the circle is the origin and its radius is $5$! $_\square$

What is the value of $k$ in the figure below?

Figure

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Since it is a circle and is touching both the $x$-axis and $y$-axis, its distance from both the axes must be the same. Since it is $3$ units away from the $x$-axis, it must be $3$ units away from the $y$-axis. Therefore,

$k = 3.\ _\square$

Another way of expressing the equation of a circle is the diametric form.

Suppose there are two points on a circle $(x_1, y_1)$ and $(x_2, y_2)$, such that they lie on the opposite ends of the same diameter, then the equation of the circle can be written as

$(x-x_1)(x-x_2) + (y-y_1)(y - y_2) = 0.$

Suppose 2 points on the circle $A= (x_1, y_1)$ and $B= (x_2, y_2)$ are diametrically opposite, then for any point $C= (x, y)$ on the circle, $\triangle ABC$ will be a right triangle, with right angle at $C$.

$\begin{aligned} AC &\perp BC\\ (m_{AC}) \cdot (m_{BC}) &= -1\\ \left(\dfrac{y - y_1}{x - x_1}\right) \cdot \left( \dfrac{y - y_2}{x - x_2}\right)= -1. \end{aligned}$

Since $x$ can be equal to $x_1$ and $x_2$,

$\begin{aligned} (y-y_1)(y - y_2 ) &= - (x - x_1) (x- x_2)\\ (x-x_1)(x-x_2) + (y-y_1)(y - y_2) &= 0.\ _\square \end{aligned}$

Find the equation of the smallest possible circle that passes through the points $(2,6)$ and $(-4, 3).$

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The circle would be the smallest if the two points were to be the endpoints of a diameter of the circle. We can use the diametric form to get

$\begin{aligned} (x - 2)(x - (-4)) + (y - 6)(y - 3) &=0\\ (x - 2)(x +4) + (y - 6)(y - 3) &=0. \ _\square \end{aligned}$