Problem 55 - College Algebra

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

Answer

The equation in standard form is \((x - 5)^{2} + (y - 3)^{2} = 64\). The center of the circle is at (5, 3) and the radius is 8.

Step 1: Rewrite the Equation

First, rearrange the equation and group \(x\) and \(y\) terms together: \((x^{2}-10x) + (y^{2}-6y) = 30\)

Step 2: Complete the Square for x and y

Complete the square for \(x\). Divide the coefficient of \(x\) by 2 and square it, which is \(\left(\frac{-10}{2}\right)^{2} = 25\).Add this to both sides: \((x^{2}-10x + 25) + (y^{2}-6y) = 30 + 25\).Repeat the same for \(y\). The coefficient of \(y\) is -6; divide -6 by 2 and square the result, \(\left(\frac{-6}{2}\right)^{2} = 9\).Add 9 to both sides: \((x^{2}-10x + 25) + (y^{2}-6y + 9) = 30 + 25 + 9\).

Step 3: Simplify the equation and write it in standard form.

Simplify the equation to get it into standard form. The standard form of a circle is \((x - h)^{2} + (y - k)^{2} = r^{2}\).\((x - 5)^{2} + (y - 3)^{2} = 64\).

Step 4: Identify the Center and the Radius

The center of the circle is at \((h, k)\), and the radius is \(r\).In our equation, \(h = 5\), \(k = 3\), and \(r^{2} = 64\). Therefore our center of the circle is at (5, 3) and radius is \( \sqrt{64} = 8 \).

Step 5: Graph the Circle

To graph, simply plot the center at (5, 3) and draw a circle with radius 8 units around this center point.