To find the points of intersection of the two given equations, we will solve the system of equations:

1. $x^2 + y^2 - 4x - 6y - 12 = 0$
2. $x^2 + y^2 - 8x - 2y - 19 = 0$

Step 1: Rewrite Equations in Standard Form

First, let's rewrite each equation by completing the square for both $x$ and $y$.

Equation 1:

1. $x^2 + y^2 - 4x - 6y - 12 = 0$

To complete the square:

• For $x$: $x^2 - 4x$
  Complete the square:
  $x^2 - 4x = (x - 2)^2 - 4$

• For $y$: $y^2 - 6y$
  Complete the square:
  $y^2 - 6y = (y - 3)^2 - 9$

Substitute back into the equation:

$(x - 2)^2 - 4 + (y - 3)^2 - 9 - 12 = 0$

Simplify:

$(x - 2)^2 + (y - 3)^2 = 25$

This is a circle centered at $(2, 3)$ with a radius of 5.

Equation 2:

2. $x^2 + y^2 - 8x - 2y - 19 = 0$

To complete the square:

• For $x$: $x^2 - 8x$
  Complete the square:
  $x^2 - 8x = (x - 4)^2 - 16$

• For $y$: $y^2 - 2y$
  Complete the square:
  $y^2 - 2y = (y - 1)^2 - 1$

Substitute back into the equation:

$(x - 4)^2 - 16 + (y - 1)^2 - 1 - 19 = 0$

Simplify:

$(x - 4)^2 + (y - 1)^2 = 36$

This is a circle centered at $(4, 1)$ with a radius of 6.

Step 2: Find Points of Intersection

Now, let's solve for the points of intersection of these two circles.

Subtract the first equation from the second:

[ [(x - 4)^2 + (y - 1)^2] - [(x - 2)^2 + (y - 3)^2] = 36 - 25 ]

Expand both squared terms:

[ [(x^2 - 8x + 16) + (y^2 - 2y + 1)] - [(x^2 - 4x + 4) + (y^2 - 6y + 9)] = 11 ]

Simplify:

$(x^2 - 8x + 16 + y^2 - 2y + 1) - (x^2 - 4x + 4 + y^2 - 6y + 9) = 11$

$x^2 - 8x + 16 + y^2 - 2y + 1 - x^2 + 4x - 4 - y^2 + 6y - 9 = 11$

Combine like terms:

$-4x + 4y + 4 = 11$

$-4x + 4y = 7$

$-x + y = \frac{7}{4}$

Step 3: Substitute and Solve for $x$ and $y$

Now substitute $y = x + \frac{7}{4}$ into either original equation. Let's use the first:

$x^2 + \left(x + \frac{7}{4}\right)^2 - 4x - 6\left(x + \frac{7}{4}\right) - 12 = 0$

Expand and simplify:

$x^2 + x^2 + \frac{14}{4}x + \frac{49}{16} - 4x - 6x - \frac{42}{4} - 12 = 0$

$2x^2 - \frac{13}{2}x + \frac{1}{16} = 0$

Multiply everything by 16 to clear the fractions:

$32x^2 - 104x + 1 = 0$

Solve this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 32$, $b = -104$, and $c = 1$:

$x = \frac{-(-104) \pm \sqrt{(-104)^2 - 4 \times 32 \times 1}}{2 \times 32}$

$x = \frac{104 \pm \sqrt{10816 - 128}}{64}$

$x = \frac{104 \pm \sqrt{10688}}{64}$

Calculate:

$x = \frac{104 \pm 103.36}{64}$

$x_1 = \frac{207.36}{64} = 3.24$

$x_2 = \frac{0.64}{64} = 0.01$

Now find corresponding $y$ values:

For $x = 3.24$, $y = 3.24 + \frac{7}{4} = 5.99$.

For $x = 0.01$, $y = 0.01 + \frac{7}{4} = 1.76$.

Points of Intersection

The points of intersection are approximately:

1. $(3.24, 5.99)$
2. $(0.01, 1.76)$

Would you like more details, or do you have any other questions?

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Related Questions:

1. How do you find the intersection points of two general circles?
2. What is the geometric interpretation of subtracting two circle equations?
3. How does completing the square help in solving circle equations?
4. How would you solve these equations graphically?
5. What are the implications if the circles do not intersect?

Tip: When working with circles, converting equations into standard form using completing the square is a powerful technique for simplifying and solving problems.