Question

find the center and the radius of the circle with the given equation. then draw the graph.
$x^2 + y^2 - 10x + 6y = 47$

use the graphing tool to graph the circle.

graph with click to enlarge graph text

Explanation:

Step1: Rewrite the equation in standard form

The general equation of a circle is \(x^{2}+y^{2}+Dx + Ey+F = 0\), and the standard form is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius. We complete the square for \(x\) and \(y\) terms.

Given equation: \(x^{2}+y^{2}-10x + 6y=47\)

Group \(x\) terms and \(y\) terms: \((x^{2}-10x)+(y^{2}+6y)=47\)

Complete the square for \(x\): The coefficient of \(x\) is \(- 10\), half of it is \(-5\), and squaring it gives \((-5)^{2}=25\).

Complete the square for \(y\): The coefficient of \(y\) is \(6\), half of it is \(3\), and squaring it gives \(3^{2} = 9\).

Add these squares to both sides of the equation:

\((x^{2}-10x + 25)+(y^{2}+6y + 9)=47+25 + 9\)

Step2: Simplify the equation

Simplify the left - hand side using the perfect square formula \((a - b)^{2}=a^{2}-2ab + b^{2}\) and \((a + b)^{2}=a^{2}+2ab + b^{2}\)

\((x - 5)^{2}+(y + 3)^{2}=81\)

Since \(81=9^{2}\), comparing with the standard form \((x - h)^{2}+(y - k)^{2}=r^{2}\), we have \(h = 5\), \(k=-3\) and \(r = 9\)

Answer:

The center of the circle is \((5,-3)\) and the radius is \(9\)