Problem 55
Question
complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$ x^{2}+y^{2}-10 x-6 y-30=0 $$
Step-by-Step Solution
Verified Answer
The center of the circle is at (5,3) and the radius of the circle is 2 units.
1Step 1: Rearranging terms grouped by variable
Start by rearranging terms, group \(x\)-terms and \(y\)-terms together: \(x^{2}-10x+y^{2}-6y=-30\)
2Step 2: Completing the square for x-terms and y-terms
We need to complete the square for \(x\)-terms and \(y\)-terms. To do this, take half of the coefficient of \(x\) \((-10/2 = -5)\) and square it \((-5)^{2}=25\). Do the same for \(y\) \((-6/2=-3)\), square it \((-3)^{2}=9\). Add these new terms in the equation on both sides: \((x^{2}-10x+25)+(y^{2}-6y+9)=-30+25+9\).\nThis simplifies to: \((x-5)^{2}+(y-3)^{2}=4\).
3Step 3: Finding the center and radius
From the equation \((x-5)^{2}+(y-3)^{2}=4\), the center of the circle is given by \((h, k) = (5, 3)\) and the radius r is given by \(\sqrt{4} = 2\).
4Step 4: Graphing the equation
Finally, plot the center point on the coordinate plane and draw a circle with a radius of 2 units from the center point. Make sure the circle intersects with the coordinate plane at points (5+2,3), (5-2,3), (5,3+2) and (5,3-2).
Key Concepts
Standard Form of a CircleCenter and Radius of a CircleGraphing Equations
Standard Form of a Circle
Understanding the standard form of a circle's equation is crucial to studying its properties and graphing it accurately. The standard form is expressed as \((x-h)^{2} + (y-k)^{2} = r^{2}\) where \((h, k)\) represents the circle's center and \(r\) is the radius. This form is derived from the Pythagorean Theorem, revealing the relationship between any point on the circle and its center.
In our exercise, the task was to rewrite the given equation into this standard form. Starting from \(x^{2}+y^{2}-10x-6y-30=0\), we grouped similar terms and completed the square, which involves adding appropriate constants to create perfect square trinomials, which eventually led to \((x-5)^{2}+(y-3)^{2}=4\), the standard form that easily shows the circle’s center and radius.
In our exercise, the task was to rewrite the given equation into this standard form. Starting from \(x^{2}+y^{2}-10x-6y-30=0\), we grouped similar terms and completed the square, which involves adding appropriate constants to create perfect square trinomials, which eventually led to \((x-5)^{2}+(y-3)^{2}=4\), the standard form that easily shows the circle’s center and radius.
Center and Radius of a Circle
To find the center and radius of a circle from its standard form equation, identify the values of \(h\), \(k\), and \(r\) in \((x-h)^{2} + (y-k)^{2} = r^{2}\). The center \((h, k)\) is the point from which all points on the circle are equidistant, and \(r\) is that constant distance, known as the radius.
In our example, after completing the square, we obtained \((x-5)^{2}+(y-3)^{2}=4\). Comparing this to the standard form, we deduce that \(h=5\) and \(k=3\), making the center \((5, 3)\). For the radius, we take the square root of \(4\), yielding \(r=2\). This process simplifies determining the circle's geometric properties without the need for extensive calculations.
In our example, after completing the square, we obtained \((x-5)^{2}+(y-3)^{2}=4\). Comparing this to the standard form, we deduce that \(h=5\) and \(k=3\), making the center \((5, 3)\). For the radius, we take the square root of \(4\), yielding \(r=2\). This process simplifies determining the circle's geometric properties without the need for extensive calculations.
Graphing Equations
Graphing the equation of a circle involves plotting the center on a coordinate plane and drawing a circle with the appropriate radius. Using the radius measurement, you can mark points on the graph that are radius-length away from the center in all directions—ensuring a perfect circle shape when connected.
From the completed square form of our equation \((x-5)^{2}+(y-3)^{2}=4\), we know the center is at \((5,3)\) and the radius is \(2\). On a graph, you start at point \((5, 3)\) and mark points at intervals of 2 units left, right, above, and below to represent the circle's edge. This method works for any circle once you've identified the center and radius from its equation.
From the completed square form of our equation \((x-5)^{2}+(y-3)^{2}=4\), we know the center is at \((5,3)\) and the radius is \(2\). On a graph, you start at point \((5, 3)\) and mark points at intervals of 2 units left, right, above, and below to represent the circle's edge. This method works for any circle once you've identified the center and radius from its equation.
Other exercises in this chapter
Problem 54
Begin by graphing the standard quadratic function, \(f(x)=x^{2} .\) Then use transformations of this graph to graph the given function. $$g(x)=x^{2}-1$$
View solution Problem 54
Graph each equation in a rectangular coordinate system. $$x=0$$
View solution Problem 55
find and simplify the difference quotient $$ \frac{f(x+h)-f(x)}{h}, h \neq 0 $$ for the given function. $$ f(x)=4 x $$
View solution Problem 55
Find a. \((f \circ g)(x) \quad \) b. \((g \circ f)(x) \quad \) c. \((f \circ g)(2) \quad \) d. \((g \circ f)(2)\) $$f(x)=4 x-3, g(x)=5 x^{2}-2$$
View solution