Problem 51 - 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 centre of the circle is at point (5, 3) and radius is of length 8 units. The standard form of the given equation of circle is \((x - 5)^2 + (y - 3)^2 = 64\)

Step 1: Rewrite the Equation

Given equation is \(x^{2}+y^{2}-10 x-6 y-30=0\). To convert the given equation into the standard form, group the 'x' terms, the 'y' terms and keep the constant term in one side. This can be rewritten as: \(x^{2}-10x + y^{2}-6y = 30\)

Step 2: Complete the squares for x and y

In the equation \(x^{2}-10x + y^{2}-6y = 30\), the square completion for x will be \((x - 5)^2\) and for y it will be \((y - 3)^2\). To prove that, one needs to add \((10/2)^2 = 25\) and \((6/2)^2 = 9\) to both sides of equation correspondingly to derive \((x - 5)^2 + (y - 3)^2 = 30 + 25 + 9\).

Step 3: Calculate the radius and Simplify the Equation

The equation \((x - 5)^2 + (y - 3)^2 = 30 + 25 + 9\) further simplifies to \((x - 5)^2 + (y - 3)^2 = 64\). This represents a circle with center at (5, 3) and the radius is square root of 64, which is 8.

Step 4: Graph the Equation

Plot the centre of the circle at point (5, 3). From the center, draw a circle with radius of length 8 units. The circle (x−5)²+(y−3)²=8 represents the equation in the graph.