Question

Write the standard form of the equation and the general form of the equation of each circle of radius \(r\) and center \((h, k)\). Graph each circle. $$ r=5 ;(h, k)=(4,-3) $$

Short answer

Standard form: \[ (x - 4)^2 + (y + 3)^2 = 25 \], General form: \[ x^2 + y^2 - 8x + 6y = 0 \]

Step-by-step solution

Identify the standard form of the equation of a circle

The standard form of the equation for a circle with radius \(r\) and center \((h, k)\) is given by: \[ (x - h)^2 + (y - k)^2 = r^2 \] In this case, the radius \(r = 5\) and the center \((h, k) = (4, -3)\).

Substitute the values into the standard form

Substitute \(r = 5\), \(h = 4\), and \(k = -3\) into the standard form equation:\[ (x - 4)^2 + (y + 3)^2 = 5^2 \]

Simplify the equation

Simplify the right-hand side of the equation by calculating \(5^2\):\[ (x - 4)^2 + (y + 3)^2 = 25 \]This is the standard form of the equation of the circle.

Convert to the general form of the equation

Expand the standard form equation to get the general form \(Ax^2 + By^2 + Cx + Dy + E = 0\):\[ (x - 4)^2 + (y + 3)^2 = 25\]Expanding:\[ x^2 - 8x + 16 + y^2 + 6y + 9 = 25 \]Combine like terms:\[ x^2 + y^2 - 8x + 6y + 25 = 25 \]Subtract 25 from both sides:\[ x^2 + y^2 - 8x + 6y = 0 \]This is the general form of the equation.

Graph the circle

Plot the center of the circle at \((4, -3)\) on a coordinate plane. Draw a circle with a radius of 5 units around the center, ensuring all points on the circle are 5 units away from the center.

Key concepts

Understanding the Standard Form of a Circle's Equation

The standard form of the equation for a circle is quite straightforward. It tells us that any point \((x, y)\) on the circle is at a fixed distance from the center \((h, k)\). This fixed distance is the radius \(r\). The standard form equation is: \[ (x - h)^2 + (y - k)^2 = r^2 \] Here, you can see three key parts:

• \(h\): the x-coordinate of the center
• \(k\): the y-coordinate of the center
• \(r\): the radius of the circle

Once you plug in the values given in the problem, it becomes easier to visualize and understand the circle. For the given problem where \((h, k) = (4, -3)\) and \(r = 5\), the equation becomes: \[ (x - 4)^2 + (y + 3)^2 = 25 \] This tells us that every point on this circle is exactly 5 units away from \((4, -3)\).
If you expand each step, you can see how everything fits together nicely.

Translating to the General Form

The general form of a circle's equation looks different from the standard form but represents the same relationship. The general form is written as: \[ Ax^2 + By^2 + Cx + Dy + E = 0 \] In our problem, we start with the standard form: \[ (x - 4)^2 + (y + 3)^2 = 25 \] We need to expand and simplify it to match the general form.
Expanding each squared term: \[ (x - 4)^2 = x^2 - 8x + 16 \] \[ (y + 3)^2 = y^2 + 6y + 9 \] Combining these, we get: \[ x^2 - 8x + 16 + y^2 + 6y + 9 = 25 \] Now, we combine like terms and subtract 25 from both sides: \[ x^2 + y^2 - 8x + 6y + 25 = 25 \] \[ x^2 + y^2 - 8x + 6y = 0 \] Now we have our general form: \[ x^2 + y^2 - 8x + 6y = 0 \] Notice how it looks more complex, but it still represents the relationship between any point \((x, y)\) and the center and radius of the circle.

Finding the Radius and Center

Understanding the radius and center of the circle is crucial as they determine the circle's position and size.

• The center \((h, k)\) is simply a point around which the entire circle is drawn. For our problem, it is \((4, -3)\).
• The radius \(r\) is the distance from the center to any point on the circle. Here, it is 5 units.

When given these values, plotting the circle becomes easier. Start by plotting the center on a graph. For \((4, -3)\), go 4 units to the right and 3 units down from the origin.
From there, measure 5 units in every direction to ensure an accurate circle. Graphing helps solidify understanding, making abstract equations more concrete.