What is the equation of the circle passing through 
(
12
,
1
)
(12,1) and 
(
2
,
−
3
)
(2,−3) with center on the line 
2
x
−
5
y
+
10
=
0
2x−5y+10=0?

To find the equation of the circle that passes through the points \( (12, 1) \) and \( (2, -3) \) with its center on the line \( 2x - 5y + 10 = 0 \), we can follow these steps:

### Step 1: General Equation of the Circle

The general equation of a circle can be expressed as:

\[
(x - h)^2 + (y - k)^2 = r^2
\]

where \( (h, k) \) is the center and \( r \) is the radius.

### Step 2: Center on the Given Line

The center \( (h, k) \) must satisfy the line equation \( 2h - 5k + 10 = 0 \). We can express \( k \) in terms of \( h \):

\[
k = \frac{2h + 10}{5}
\]

### Step 3: Distance from the Center to the Points

The circle passes through the points \( (12, 1) \) and \( (2, -3) \). Therefore, the distances from the center to these points should be equal to the radius \( r \):

1. From \( (h, k) \) to \( (12, 1) \):
   \[
   r^2 = (h - 12)^2 + (k - 1)^2
   \]

2. From \( (h, k) \) to \( (2, -3) \):
   \[
   r^2 = (h - 2)^2 + (k + 3)^2
   \]

### Step 4: Setting the Equations Equal

Setting the two equations for \( r^2 \) equal gives:

\[
(h - 12)^2 + (k - 1)^2 = (h - 2)^2 + (k + 3)^2
\]

Expanding both sides:

\[
(h^2 - 24h + 144) + (k^2 - 2k + 1) = (h^2 - 4h + 4) + (k^2 + 6k + 9)
\]

### Step 5: Simplifying the Equation

Cancelling \( h^2 \) and \( k^2 \) from both sides results in:

\[
-24h + 144 - 2k + 1 = -4h + 4 + 6k + 9
\]

Simplifying further:

\[
-20h + 132 - 8k = 0
\]

Rearranging gives:

\[
20h + 8k = 132 \quad \Rightarrow \quad 5h + 2k = 33
\]

### Step 6: Substitute for \( k \)

Now substitute \( k = \frac{2h + 10}{5} \) into the equation \( 5h + 2k = 33 \):

\[
5h + 2\left(\frac{2h + 10}{5}\right) = 33
\]

Multiply through by 5 to eliminate the fraction:

\[
25h + 2(2h + 10) = 165
\]

Expanding gives:

\[
25h + 4h + 20 = 165 \quad \Rightarrow \quad 29h = 145 \quad \Rightarrow \quad h = 5
\]

### Step 7: Find \( k \)

Now substitute \( h = 5 \) back to find \( k \):

\[
k = \frac{2(5) + 10}{5} = \frac{20}{5} = 4
\]

### Step 8: Find the Radius

Now we can find \( r^2 \) using either point. Using point \( (12, 1) \):

\[
r^2 = (5 - 12)^2 + (4 - 1)^2 = (-7)^2 + (3)^2 = 49 + 9 = 58
\]

### Step 9: Final Equation of the Circle

The equation of the circle is:

\[
(x - 5)^2 + (y - 4)^2 = 58
\]

This can be expanded to:

\[
x^2 - 10x + 25 + y^2 - 8y + 16 = 58
\]

Simplifying gives:

\[
x^2 + y^2 - 10x - 8y - 17 = 0
\]

### Final Result

Thus, the equation of the circle is:

\[
x^2 + y^2 - 10x - 8y - 17 = 0
\]

---

Would you like further details or explanations on any part of this process? Here are some related questions you might find interesting:

1. How do you derive the radius from the center and a point on the circle?
2. What methods can be used to find the center of a circle given points?
3. How do you convert the standard form of a circle into general form?
4. Can you explain the geometric interpretation of a circle's center and radius?
5. What are the characteristics of circles that intersect two specific points? 

**Tip:** Always check your calculations by substituting the points back into the final equation to ensure they satisfy it.

What is the equation of the circle passing through (12, 1) and (2, -3) with center on the line 2x - 5y + 10 = 0?

Discover how to find the equation of a circle that passes through the points (12, 1) and (2, -3) with its center constrained to the line 2x - 5y + 10 = 0. This solution covers essential geometric concepts and step-by-step calculations suitable for high school students.

Math Problem Statement

What is the equation of the circle passing through ( 12 , 1 ) (12,1) and ( 2 , − 3 ) (2,−3) with center on the line 2 x − 5 y + 10

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