Solving (x-h)^2+(y-k)^2=r^2: Reverse + to - & Square? (2024)

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Related to Solving (x-h)^2+(y-k)^2=r^2: Reverse + to - & Square? 1. What does (x-h)^2+(y-k)^2=r^2 represent? 2. How do I solve (x-h)^2+(y-k)^2=r^2 for x and y? 3. Can (x-h)^2+(y-k)^2=r^2 be rewritten as a standard form equation? 4. How do I convert from + to - in the equation (x-h)^2+(y-k)^2=r^2? 5. How do I use (x-h)^2+(y-k)^2=r^2 to graph a circle? Similar threads
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In summary, to solve this particular problem, you need to identify the center of the given circle, which is represented by the parameters $h$ and $k$, and then use the distance formula and Pythagorean theorem to find the radius and write the equation of the circle in standard form. The use of $h$ and $k$ may seem confusing, but they are just arbitrary parameters for the center of the circle.

  • #1

OMGMathPLS

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View attachment 3289

So... you just reverse the - and positive sign and square the ending =? Is that how you solve this?

Because writing it out does not make sense. How do you go from positive to turning it into a negative?

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  • #3

OMGMathPLS

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The h,k are the relation, the origin co-ordinates, right? The center of the circle?

And this is part of the Pythagorean theorem but we are just adding in the x and y coordinates to shift it, right?

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  • #4

MarkFL

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MHB

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OMGMathPLS said:

The h,k are the relation, the origin co-ordinates, right? The center of the circle?

And this is part of the Pythagorean theorem but we are just adding in the x and y coordinates to shift it, right?

The way I look at it, a circle is defined to be the locus of all points a given distance, the radius $r$, from some central point, the focus or center. If we let the center be at $(h,k)$, and an arbitrary point on the circle be $(x,y)$, then the distance formula gives us:

\(\displaystyle \sqrt{(x-h)^2+(y-k)^2}=r\)

And then squaring, we obtain:

\(\displaystyle (x-h)^2+(y-k)^2=r^2\)

  • #5

OMGMathPLS

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So if it's actually a positive we always have to keep it as a negative by turning it into a -(-) .

Do you know what its an h and k? This is really confusing me. In other countries it's called an a and b right? I found when I think of hong kong the hong is the x and the kong is the y it helps but it just makes no sense why it's h,k especially if that was arbitrary.

Thanks for your answer.

  • #6

MarkFL

Gold Member

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Yes, the use of $h$ and $k$ in this context are just an arbitrary convention used here in the U.S., and perhaps in other places as well. You can use any parameters for the center that you want, such as $(a,b)$.

Related to Solving (x-h)^2+(y-k)^2=r^2: Reverse + to - & Square?

1. What does (x-h)^2+(y-k)^2=r^2 represent?

(x-h)^2+(y-k)^2=r^2 is the equation for a circle with center (h,k) and radius r. The variables x and y represent the coordinates of any point on the circle.

2. How do I solve (x-h)^2+(y-k)^2=r^2 for x and y?

To solve this equation, you will need to use the reverse and square method. First, subtract k from both sides of the equation. Then, take the square root of both sides. Finally, add h to both sides to solve for x. Repeat the process to solve for y.

3. Can (x-h)^2+(y-k)^2=r^2 be rewritten as a standard form equation?

Yes, (x-h)^2+(y-k)^2=r^2 can be rewritten as (x-h)^2+(y-k)^2=r^2 by expanding the squared terms and grouping like terms together.

4. How do I convert from + to - in the equation (x-h)^2+(y-k)^2=r^2?

To convert from + to - in the equation (x-h)^2+(y-k)^2=r^2, simply change the sign of h and k. For example, if the equation is (x+3)^2+(y+4)^2=r^2, the converted equation would be (x-3)^2+(y-4)^2=r^2.

5. How do I use (x-h)^2+(y-k)^2=r^2 to graph a circle?

To graph a circle using (x-h)^2+(y-k)^2=r^2, first plot the point (h,k) on a coordinate plane. This represents the center of the circle. Then, use the radius r to plot points on the circle. For example, if r=3, you would plot points (h+3,k), (h-3,k), (h,k+3), (h,k-3) on the coordinate plane. Finally, connect these points to form a circle.

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