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Author: nsharp1
No. Questions: 10
Created: Jan 2, 2019

# Derive the Equation of a Circle

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Derive the equation for a circle with center at (3,4) and a radius of 2.
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.1
B.
If there were multiple correct answers in the previous question, will this result in different triangles?
1. No, they will be congruent (one will be a transformation of the other).
2. Yes, one will be larger than the other.
3. Yes, one will be an isosceles right triangle, the other will be a scalene right triangle.
4. No, there was only one right answer.
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.1
E.
Now that we have the lengths of the legs of the triangle, what theorem, property, or formula correctly relates them and why is it used?
1. The formula for the area of a triangle, $A = 1/2 (y-3) * (x-4)$, since we know the height and base of the triangle.
2. Heron's Formula, $A = sqrt(s*(s-a)*(s-b)*(s-c)), " where " s = (2 + (y-4)^2 + (x-3)^2 )/2$, because the length of all three sides is known.
3. Pythagorean Theorem, $(x-3)^2 + (y-4)^2 = 4$, since the lengths of the three sides of the right triangle are known, and this introduces no extra variables.
4. Law of Cosines, $|x-3|^2 = 4 + |y-4|^2 - 2*2*|y-4|*cos90°$, because the lengths of the three sides of the triangle, and the measure of the right angle, are known, and this introduces no extra variables.
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.1
F.
Having found the equation of the circle centered at $(3,4)$ with radius of length 2 in the previous question, which of the following best describes the meaning of this equation?
1. There are always two points inside a circle with which to form a right triangle using the radius of the circle as the hypotenuse.
2. Any point (x,y) that is 2 units away from the center of this circle lies on the circle.
3. There exists one point, (x,y), which solves this equation.
4. All points that lie on this circle also must be part of a right triangle.
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.1
G.
In the first question, the stipulation on the point $(x,y)$ was made that the x-coordinate could not be 3, and the y-coordinate could not be 4. Why was this?
1. These points do not actually occur on the circle, and so would have caused an error in the derivation.
2. If a point met either of these criteria, it would be either directly above, below, to the right of, or to the left of the center of the circle. This would have made creating a right triangle with legs that have slope of 0 and undefined impossible.
3. These points would have resulted in complex values (calculations involving imaginary numbers), and led to serious complications in the derivation process.
4. The points that meet these criteria result in triangles which are not right triangles.
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.1
H.
For the points which are excluded by the condition mentioned in the previous question, does the equation of the circle not work for them? Why or why not?
1. It doesn't, because either the x or y distance becomes zero meaning an entirely different equation is required.
2. It doesn't, these points are always excluded, but just usually ignored.
3. It does, however these points would just have introduced complex numbers into the derivation and so were avoided.
4. It does, these points are simply special cases where the distance from the center to the circle is all in either the x or y direction.
Grade 11 Nonlinear Equations and Functions CCSS: HSG-GPE.A.1
I.
In the first question, a right triangle was created, using the radius of the circle as the hypotenuse, such that one leg had a slope of zero, and the other leg's slope was undefined. Why was this?
1. This is the only right triangle that can be formed for the given hypotenuse.
2. This is the only right triangle that can be formed for the given hypotenuse that is completely inside the given circle.
3. Using a right triangle with legs that have either a slope of zero or an undefined slope means that they represent the vertical and horizontal distance of the hypotenuse.
4. There is no need for this triangle, but it makes the question more difficult because it results in two possible correct answers.