Question

The diameter of Circle Q terminates on the circumference of the circle at (3,0) and (3,-7).in standard form. Show all of your work for full credit.

Answer 1

To find the standard form of the equation that represents this circle, we proceed as follows:

Step 1: Recall the standard form of the equation of any circle as:

`(x-a)^2+(y-b)^2=r^2`

Where:

(a, b) == the coordinates of the center of the circle

r == the radius of the circle

Step 2: Find the coordinates of the center of the circle, given the endpoints of the diameter

To find the coordinate of the center of the circle, we simply find the midpoint of the line that connects the two points together.

From coordinate geometry, if we have any two points (x1, y1) and (x2, y2) the coordinate of the midpoint of the line that connects the two points together is as follows:

`\text{midpoint}=\text{ (}(x_1+x_2)/(2),(y_1+y_2)/(2))`

Given that (x1, y1) == (3, 0), and (x2, y2) == (3, -7), the midpoint is:

`\begin{gathered} \text{midpoint}=\text{ (}(x_1+x_2)/(2),(y_1+y_2)/(2)) \\ \text{midpoint}=\text{ (}\frac{3_{}+3_{}}{2},\frac{0_{}+(-7)}{2}) \\ \text{midpoint}=\text{ (}\frac{6_{}}{2},\frac{-7_{}}{2}) \\ \text{midpoint}=\text{ (}3,\frac{-7_{}}{2}) \end{gathered}`

Therefore, the coordinates of the center of the circle is (3, -7/2)

Step 3: Find the value of the radius of the circle

The value of the radius of the circle will be the distance between the center of the circle and any of the endpoints of the diameter.

From coordinate geometry, if we have any two points (x1, y1) and (x2, y2), the distance between them is given as:

`\begin{gathered} d^2=(x_2-x_1)^2+(y_2-y_1)^2 \\ d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \end{gathered}`

Let the center of the circle be (x1, y1) == (3, -7/2), and (x2, y2) == (3, 0), thus the distance is:

`\begin{gathered} d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ d=\sqrt[]{(3-3)^2+(0-(-(7)/(2)))^2} \\ d=\sqrt[]{(0)^2+(0+(7)/(2))^2} \\ d=\sqrt[]{0+((7)/(2))^2} \\ d=\sqrt[]{((7)/(2))^2} \\ d=(7)/(2)^{} \end{gathered}`

Therefore, the radius of the circle is 7/2

Step 4: Now, with the coordinate of the center of the circle and its radius obtained, we can write out the equation of the circle in standard form, as follows:

`(x-a)^2+(y-b)^2=r^2`

Where:

(a, b) == the coordinates of the center of the circle

r == the radius of the circle

Now, with: (a, b) == (3, -7/2) and r = 7/2, we have:

`\begin{gathered} (x-a)^2+(y-b)^2=r^2 \\ (x-3)^2+(y-(-(7)/(2)))^2=((7)/(2))^2 \\ (x-3)^2+(y+(7)/(2))^2=(7^2)/(2^2)^{} \\ (x-3)^2+(y+(7)/(2))^2=\frac{49^{}}{4}^{} \end{gathered}`

Thus, as shown above, the equation of the circle in its standard form is as follows:

(x - 3)^2 + (y + 7/2)^2 = 49/2