The equation of a circle in standard form is given by:

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

where $(h, k)$ is the center of the circle and $r$ is its radius.

Step 1: Identify the center

The center of the circle is given as $(-3, 4)$, so: $h = -3 \quad \text{and} \quad k = 4$

Step 2: Find the radius

The radius $r$ is the distance between the center $(-3, 4)$ and the point on the circle $(6, 2)$. The distance formula is: $r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Substituting the values: $r = \sqrt{(6 - (-3))^2 + (2 - 4)^2}$ $r = \sqrt{(6 + 3)^2 + (-2)^2}$ $r = \sqrt{9^2 + (-2)^2}$ $r = \sqrt{81 + 4}$ $r = \sqrt{85}$

Step 3: Write the equation

Now substitute $(-3, 4)$ for $(h, k)$ and $\sqrt{85}$ for $r$ into the standard form equation: $(x - (-3))^2 + (y - 4)^2 = (\sqrt{85})^2$ Simplify: $(x + 3)^2 + (y - 4)^2 = 85$

Final Answer:

The equation of the circle in standard form is: $(x + 3)^2 + (y - 4)^2 = 85$

Let me know if you'd like further clarification or additional examples.

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Here are 5 related questions you might find useful:

1. How would you write the general form of this circle's equation?
2. What is the radius of the circle in decimal form?
3. How do you determine if another point lies on this circle?
4. How does the standard form equation change if the center is shifted?
5. Can you derive the equation of a circle using the midpoint and radius formula?

Tip: Always double-check calculations for the radius using the distance formula to avoid errors in squaring or simplifications.