The standard equation of a circle with center (h, k) and radius r is given by (x - h)² + (y - k)² = r². This formula is derived from the distance formula, where every point (x, y) on the circle is at a fixed distance r from the center (h, k).

The standard equation of a circle with center (h, k) and radius r is given by (x - h)² + (y - k)² = r². This formula is derived from the distance formula, where every point (x, y) on the circle is at a fixed distance r from the center (h, k).

Determine the equation of the circle with radius 4 and Centre (-2, 3).

Answer: The equation of the circle is $(x + 2)^2 + (y - 3)^2 = 16$ .

Step-by-step solution

Step 1: Recall the standard equation of a circle

The standard equation of a circle with center (h, k) and radius r is given by (x - h)^2 + (y - k)^2 = $r^2$ . This formula is derived from the distance formula, where every point (x, y) on the circle is at a fixed distance r from the center (h, k).

Step 2: Identify the given values

From the problem statement, we are given that the center of the circle is (-2, 3) and the radius is 4. Therefore, we can identify $h = -2$ , $k = 3$ $, and r = 4$ .

Step 3: Substitute the values into the equation

Now, we substitute the identified values of $h = -2$ $, k = 3$ , and $r = 4$ into the standard equation of the circle. This gives us (x - (-2))^2 + $(y - 3)^2 = 4^2$ .

Step 4: Simplify the equation

Finally, we simplify the equation. Subtracting a negative number is equivalent to adding, so (x - (-2)) becomes $(x + 2)$ . Also, $4^2$ simplifies to 16. Thus, the equation of the circle is $(x + 2)^2 + (y - 3)^2 = 16$ .

More from this chapter

All Conic Sections solutions →

Solve your own maths question free →