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given a circle where the center is (-4, -6) and another point on the ci…

Question

given a circle where the center is (-4, -6) and another point on the circle is (-1, -2), (a) write an equation of the circle in standard form. (b) graph the circle.

Explanation:

Step1: Recall the standard form of the circle equation

The standard form of a circle equation is \((x - h)^2+(y - k)^2=r^{2}\), where \((h,k)\) is the center of the circle and \(r\) is the radius.

Step2: Calculate the radius

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). Here, \((h,k)=(-4,-6)\) (center) and \((x_2,y_2)=(-1,-2)\) (a point on the circle).
\[

$$\begin{align*} r&=\sqrt{(-1-(-4))^2+(-2 - (-6))^2}\\ &=\sqrt{(-1 + 4)^2+(-2 + 6)^2}\\ &=\sqrt{3^{2}+4^{2}}\\ &=\sqrt{9 + 16}\\ &=\sqrt{25}\\ &=5 \end{align*}$$

\]

Step3: Write the equation of the circle

Substitute \(h=-4\), \(k = - 6\), and \(r = 5\) into the standard - form equation \((x - h)^2+(y - k)^2=r^{2}\).
\[

$$\begin{align*} (x-(-4))^2+(y-(-6))^2&=5^{2}\\ (x + 4)^2+(y + 6)^2&=25 \end{align*}$$

\]

Answer:

(a) \((x + 4)^2+(y + 6)^2 = 25\)
(b) To graph the circle:

  1. Locate the center \((-4,-6)\) on the coordinate - plane.
  2. Since the radius \(r = 5\) (from part (a) as \(r^{2}=25\)), from the center, move 5 units up, down, left, and right to plot four points on the circle. Then draw a smooth curve passing through these points to represent the circle.