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Chapter 2: Problem 55

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}-10 x-6 y-30=0$$

Short Answer

Expert verified
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 by step solution

01

Rewrite the Equation

First, rearrange the equation and group \(x\) and \(y\) terms together: \((x^{2}-10x) + (y^{2}-6y) = 30\)
02

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: \((x^{2}-10x + 25) + (y^{2}-6y + 9) = 30 + 25 + 9\).
03

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 \((x - h)^{2} + (y - k)^{2} = r^{2}\).\((x - 5)^{2} + (y - 3)^{2} = 64\).
04

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 \( \sqrt{64} = 8 \).
05

Graph the Circle

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

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