Question

use the pythagorean theorem to derive the equation of the circle centered at (2, -3) with a radius of 2.
to which triangle should you apply the pythagorean theorem?
two coordinate plane diagrams with circles and triangles
complete the paragraph.
the length of the triangle’s horizontal leg is ____. the length of its vertical leg is __. the length of its hypotenuse is __. by the pythagorean theorem, the equation of the circle is ____.

Explanation:

Step1: Identify the correct triangle

The circle is centered at \((2, -3)\) with radius 2. For a general point \((x, y)\) on the circle, the triangle formed should have vertices at the center \((2, -3)\), the point \((x, y)\), and the projection of \((x, y)\) onto the horizontal/vertical line through the center. The right graph (second graph) has vertices \((x, y)\), \((2, y)\), and \((2, -3)\), which is a right triangle with horizontal leg along \(y\)-constant line and vertical leg along \(x\)-constant line. So we apply Pythagorean theorem to the triangle in the right graph (second graph).

Step2: Horizontal leg length

The horizontal leg is the distance between \((x, y)\) and \((2, y)\), which is \(|x - 2|\), so length is \(|x - 2|\) (or \(x - 2\) if \(x\geq2\), but generally \(|x - 2|\)).

Step3: Vertical leg length

The vertical leg is the distance between \((2, y)\) and \((2, -3)\), which is \(|y - (-3)| = |y + 3|\), so length is \(|y + 3|\).

Step4: Hypotenuse length

The hypotenuse is the radius of the circle, which is 2 (since it's the distance from center \((2, -3)\) to point \((x, y)\) on the circle).

Step5: Derive circle equation

By Pythagorean theorem, \((\text{horizontal leg})^2 + (\text{vertical leg})^2 = (\text{hypotenuse})^2\), so \((x - 2)^2 + (y + 3)^2 = 2^2 = 4\).

Answer:

The triangle in the right graph (second graph) should be used.
Horizontal leg length: \(|x - 2|\) (or \(x - 2\) in absolute value form)
Vertical leg length: \(|y + 3|\)
Hypotenuse length: \(2\)
Circle equation: \((x - 2)^2 + (y + 3)^2 = 4\)

(For the triangle identification, if we consider the two graphs: first graph has horizontal leg \(|x - 2|\) (since \(y=-3\) for both \((2,-3)\) and \((x,-3)\)) and vertical leg \(|y + 3|\) (since \(x=2\) for \((2,-3)\) and \((2,y)\)), but actually the correct triangle is the one with vertices \((x,y)\), \((2,y)\), \((2,-3)\) (right graph). Wait, maybe I mixed up. Wait, the first graph: points \((2,-3)\), \((x,-3)\), \((2,y)\) – that's also a right triangle. Wait, let's recast:

Wait, center \((2, -3)\), point on circle \((x, y)\). The horizontal distance between \((x,y)\) and \((2,y)\) is \(|x - 2|\), vertical distance between \((2,y)\) and \((2,-3)\) is \(|y + 3|\), and hypotenuse is radius 2. So both triangles are right triangles, but the one with vertices \((x,y)\), \((2,y)\), \((2,-3)\) (right graph) or \((2,-3)\), \((x,-3)\), \((2,y)\) (left graph) – actually, both are right triangles, but the standard way is to take the triangle with legs parallel to axes from \((x,y)\) to \((h,k)\) (center \((h,k)=(2,-3)\)). So the horizontal leg is \(|x - h| = |x - 2|\), vertical leg \(|y - k| = |y - (-3)| = |y + 3|\), hypotenuse \(r = 2\). So the equation is \((x - 2)^2 + (y + 3)^2 = 2^2 = 4\).

So correcting:

Horizontal leg length: \(|x - 2|\) (or \((x - 2)\) when squared, the absolute value doesn't matter as we square it)
Vertical leg length: \(|y + 3|\) (or \((y + 3)\) when squared)
Hypotenuse length: \(2\)
Circle equation: \((x - 2)^2 + (y + 3)^2 = 4\)

And the triangle to apply Pythagorean theorem is the right triangle with vertices \((x, y)\), \((2, y)\), and \((2, -3)\) (the triangle in the right graph, second graph).)