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Chapter 1: Problem 65

Find the center and radius of the circle whose equation is given. $$x^{2}+y^{2}+6 x-4 y-15=0$$

Short Answer

Expert verified
Question: Determine the center and radius of the circle with the equation \(x^{2} + y^{2} + 6x - 4y - 15 = 0\). Answer: The center of the circle is \((-3, 2)\), and the radius is \(2\).

Step by step solution

01

Rewrite the equation in the standard form

We begin by completing the square for the x and y terms in the given equation: $$x^{2}+y^{2}+6 x-4 y-15=0.$$ Rearrange the equation to separate x and y terms: $$(x^{2}+6x)+(y^{2}-4y)=-15.$$ Complete the square for the x terms: Add \(n_{x}\) and subtract \(n_{x}\) (\(n_{x} = \frac{6}{2}^{2} = 9\)): $$x^{2}+6x + n_{x}-n_{x}+(y^{2}-4y)=-15$$ $$(x^2 + 6x + 9) - 9 + (y^2 - 4y) = -15$$. Complete the square for the y terms: Add \(n_y\) and subtract \(n_y\) (\(n_y = \frac{4}{2}^{2} = 4\)): $$(x^{2}+6x+9)+(-4y+y^{2}+n_{y}-n_{_y})=-15$$ $$(x^2 + 6x + 9) + (y^2 - 4y + 4) - 4 = -15$$. Finally, rewrite the equation in the standard form of a circle equation \((x-a)^{2}+(y-b)^{2}=r^{2}\): $$(x+3)^{2}+(y-2)^{2}=4.$$
02

Identify the center and the radius

Now that we have the equation in the standard form, we can identify the center \((a, b)\) and the radius \(r\): $$(x-(-3))^{2}+(y-2)^{2}=2^{2}$$. Comparing coefficients, we get the center \((-3, 2)\) and the radius \(2\).
03

State the answer

The center of the circle is \((-3, 2)\), and the radius is \(2\).

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