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Chapter 4: Problem 19

Find the equation of the circle in the \(x y\) -plane centered at (3,-2) with radius 7 .

Short Answer

Expert verified
The equation of the circle in the xy-plane centered at (3, -2) with radius 7 is \((x-3)^2 + (y+2)^2 = 49\).

Step by step solution

01

Write down the standard equation of a circle

The standard equation of a circle is \[ (x-a)^2 + (y-b)^2 = r^2, \]where \((a,b)\) is the center of the circle and \(r\) is the radius.
02

Substitute the given center and radius into the equation

The center of the circle is given as \((3,-2)\), so \(a = 3\) and \(b = -2\). The radius is given as 7, so \(r = 7\). We substitute these values into the standard equation: \[ (x-3)^2 + (y+2)^2 = 7^2. \]
03

Simplify the equation

Now we need to simplify the equation: \[(x-3)^2 + (y+2)^2 = 49. \]This is the equation of the circle in the xy-plane centered at (3,-2) with radius 7. The final answer is \[ (x-3)^2 + (y+2)^2 = 49. \]

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