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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.
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
\[(x - 2)^2 + (y + 3)^2 = 23 + 4 + 9 = 36. \ _\square\]
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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
\[\begin{align} x^2 + 2gx + y^2 + 2fy + c &= 0\\ (x+g)^2 + (y+f)^2 - g^2 - f^2 + c &= 0\\ (x+g)^2 + (y+f)^2 &= g^2 + f^2 - c. \end{align}\]
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.
Examples
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 \((x-3)^2+(y-5)^2=49\).
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?
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 \((1,7)\). \(_\square\)
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What are the radius and center of the circle whose equation is \(x^2-4x+y^2+2x=44\)?
We can apply completing the square method to left hand side:
\[\begin{align} x^2-4x+2^2+y^2+2x+1^2&=49+2^2+1^2\\ (x^2-4x+4)+(y^2+2x+1)&=49\\ (x-2)^2+(y+1)^2&=7^2. \end{align}\]
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 \(x^2+y^2=25\)?
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
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,
\[k = 3.\ _\square\]
Diametric Form
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 \(C\).
\[\begin{align} AC &\perp BC\\ (m_{AC}) \cdot (m_{BC}) &= -1\\ \left(\dfrac{y - y_1}{x - x_1}\right) \cdot \left( \dfrac{y - y_2}{x - x_2}\right)= -1. \end{align}\]
Since \(x\) can be equal to \(x_1\) and \(x_2\),
\[\begin{align} (y-y_1)(y - y_2 ) &= - (x - x_1) (x- x_2)\\ (x-x_1)(x-x_2) + (y-y_1)(y - y_2) &= 0.\ _\square \end{align}\]
Find the equation of the smallest possible circle that passes through the points \((2,6)\) and \((-4, 3).\)
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
\[\begin{align} (x - 2)(x - (-4)) + (y - 6)(y - 3) &=0\\ (x - 2)(x +4) + (y - 6)(y - 3) &=0. \ _\square \end{align}\]