Cirkelvergelijking door het Voltooid Vierkant

Problem

Transform the circle equation $x^2 + y^2 + 8x - 6y + 7 = 0$ into standard form, find its center and radius, and determine whether the point $(1,2)$ lies inside, on, or outside the circle.

Step 1: Complete the square

Start by grouping the $x$-terms and $y$-terms and moving the constant to the right:

$$x^2 + 8x + y^2 - 6y = -7$$

Complete the square for each variable. Half of $8$ is $4$, so add $4^2 = 16$. Half of $-6$ is $-3$, so add $(-3)^2 = 9$.

$$x^2 + 8x + 16 + y^2 - 6y + 9 = -7 + 16 + 9$$

This gives

$$(x+4)^2 + (y-3)^2 = 18$$

Step 2: Read the center and radius

From the standard form $(x-h)^2 + (y-k)^2 = r^2$, the center is $(-4,3)$ and the radius is

$$r = \sqrt{18} = 3\sqrt{2}$$

Step 3: Test the point $(1,2)$

Use the distance formula from the center $(-4,3)$ to the point $(1,2)$:

$$d = \sqrt{(1-(-4))^2 + (2-3)^2}$$ $$d = \sqrt{25 + 1} = \sqrt{26}$$

Since $\sqrt{26} > \sqrt{18}$, the point lies outside the circle.

Answer

The circle is $(x+4)^2 + (y-3)^2 = 18$, with center $(-4,3)$ and radius $3\sqrt{2}$, and the point $(1,2)$ lies outside the circle.