(x - 3)^2 + (y - 5)^2 = 10; x^2 + (y + 1)^2 = 25

We are given two equations:

1. \((x - 3)^2 + (y - 5)^2 = 10\)
2. \(x^2 + (y + 1)^2 = 25\)

These represent two circles on the \(xy\)-plane, and we can analyze them to find their points of intersection by solving for \(x\) and \(y\).

### Step 1: Expanding the equations

Let’s first expand both equations.

#### Expanding equation 1:
\[
(x - 3)^2 + (y - 5)^2 = 10
\]
Expanding each term:
\[
(x - 3)^2 = x^2 - 6x + 9
\]
\[
(y - 5)^2 = y^2 - 10y + 25
\]
Thus, equation 1 becomes:
\[
x^2 - 6x + 9 + y^2 - 10y + 25 = 10
\]
Simplifying:
\[
x^2 + y^2 - 6x - 10y + 34 = 10
\]
\[
x^2 + y^2 - 6x - 10y + 24 = 0 \tag{1}
\]

#### Expanding equation 2:
\[
x^2 + (y + 1)^2 = 25
\]
Expanding each term:
\[
(y + 1)^2 = y^2 + 2y + 1
\]
Thus, equation 2 becomes:
\[
x^2 + y^2 + 2y + 1 = 25
\]
Simplifying:
\[
x^2 + y^2 + 2y - 24 = 0 \tag{2}
\]

### Step 2: Subtract the two equations

Now subtract equation (2) from equation (1):
\[
(x^2 + y^2 - 6x - 10y + 24) - (x^2 + y^2 + 2y - 24) = 0
\]
Simplifying:
\[
-6x - 10y + 24 - 2y + 24 = 0
\]
\[
-6x - 12y + 48 = 0
\]
\[
6x + 12y = 48
\]
Dividing by 6:
\[
x + 2y = 8 \tag{3}
\]

### Step 3: Solve for \(x\) and \(y\)

We now have the linear equation \(x + 2y = 8\). We can solve for \(x\) in terms of \(y\):
\[
x = 8 - 2y
\]

### Step 4: Substitute into one of the original equations

Substitute \(x = 8 - 2y\) into equation (2):
\[
(8 - 2y)^2 + y^2 + 2y - 24 = 0
\]
Expanding \((8 - 2y)^2\):
\[
(8 - 2y)^2 = 64 - 32y + 4y^2
\]
Thus, the equation becomes:
\[
64 - 32y + 4y^2 + y^2 + 2y - 24 = 0
\]
Simplifying:
\[
5y^2 - 30y + 40 = 0
\]
Dividing by 5:
\[
y^2 - 6y + 8 = 0
\]
Factoring:
\[
(y - 4)(y - 2) = 0
\]
So, \(y = 4\) or \(y = 2\).

### Step 5: Find the corresponding \(x\)-values

For \(y = 4\):
\[
x = 8 - 2(4) = 8 - 8 = 0
\]

For \(y = 2\):
\[
x = 8 - 2(2) = 8 - 4 = 4
\]

### Final solution:
The points of intersection are:
\[
(0, 4) \quad \text{and} \quad (4, 2)
\]

Would you like more details on any of the steps?

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Here are 5 related questions:

1. How do you find the equation of a circle given its center and radius?
2. Can you solve a system of a linear equation and a quadratic equation graphically?
3. What are the key properties of circles that help in solving intersection problems?
4. How would this problem change if one of the equations represented a line instead of a circle?
5. How do you classify the intersection points of two geometric shapes?

**Tip:** When solving systems involving quadratics, try to simplify the problem by eliminating variables through substitution or elimination.

Discover how to find the points of intersection between two circles given by the equations (x - 3)^2 + (y - 5)^2 = 10 and x^2 + (y + 1)^2 = 25. This problem explores key concepts in algebra and geometry, such as expanding equations and solving systems of quadratic and linear equations. Ideal for students in Grades 10-12.

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