Point A(- 1, y) and B(5, 7) lie on a circle with centre 0(2, -3y). Find the values of y. Hence find the
radius of the circle.

To find the values of \( y \) and the radius of the circle, we can use the fact that both points \( A(-1, y) \) and \( B(5, 7) \) lie on the circle with center \( O(2, -3y) \). This means that the distances from the center to each point must be equal to the radius of the circle.

### Step 1: Set Up the Equations

#### Distance from \( O \) to \( A \)

The distance between the center \( O(2, -3y) \) and point \( A(-1, y) \) is given by the distance formula:

\[
OA = \sqrt{(2 - (-1))^2 + (-3y - y)^2}
\]

Simplifying the expression inside the square root:

\[
= \sqrt{(2 + 1)^2 + (-3y - y)^2} 
\]
\[
= \sqrt{3^2 + (-4y)^2} 
\]
\[
= \sqrt{9 + 16y^2}
\]

#### Distance from \( O \) to \( B \)

The distance between the center \( O(2, -3y) \) and point \( B(5, 7) \) is:

\[
OB = \sqrt{(2 - 5)^2 + (-3y - 7)^2}
\]

Simplifying the expression inside the square root:

\[
= \sqrt{(-3)^2 + (-3y - 7)^2}
\]
\[
= \sqrt{9 + (3y + 7)^2}
\]

### Step 2: Equate the Distances

Since \( OA = OB \), we equate the two distances:

\[
\sqrt{9 + 16y^2} = \sqrt{9 + (3y + 7)^2}
\]

Squaring both sides to eliminate the square roots:

\[
9 + 16y^2 = 9 + (3y + 7)^2
\]

### Step 3: Expand and Simplify

Expanding the right side:

\[
(3y + 7)^2 = 9y^2 + 42y + 49
\]

Substituting back into the equation:

\[
9 + 16y^2 = 9 + 9y^2 + 42y + 49
\]

Subtract 9 from both sides:

\[
16y^2 = 9y^2 + 42y + 49
\]

Subtract \( 9y^2 + 42y + 49 \) from both sides:

\[
16y^2 - 9y^2 - 42y - 49 = 0
\]

This simplifies to:

\[
7y^2 - 42y - 49 = 0
\]

### Step 4: Solve the Quadratic Equation

The quadratic equation is:

\[
7y^2 - 42y - 49 = 0
\]

Divide the whole equation by 7:

\[
y^2 - 6y - 7 = 0
\]

We can solve this using the quadratic formula, where \( a = 1 \), \( b = -6 \), and \( c = -7 \):

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substituting the values:

\[
y = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times (-7)}}{2 \times 1}
\]
\[
y = \frac{6 \pm \sqrt{36 + 28}}{2}
\]
\[
y = \frac{6 \pm \sqrt{64}}{2}
\]
\[
y = \frac{6 \pm 8}{2}
\]

Thus, we have two solutions:

\[
y = \frac{6 + 8}{2} = 7
\]
\[
y = \frac{6 - 8}{2} = -1
\]

### Step 5: Find the Radius

We can use either value of \( y \) to find the radius. Let's use \( y = 7 \):

The center of the circle is \( O(2, -3 \times 7) = O(2, -21) \).

Using point \( B(5, 7) \) to find the radius \( r \):

\[
r = OB = \sqrt{(2 - 5)^2 + (-21 - 7)^2}
\]
\[
= \sqrt{(-3)^2 + (-28)^2}
\]
\[
= \sqrt{9 + 784}
\]
\[
= \sqrt{793}
\]

Therefore, the radius is \(\sqrt{793}\).

### Summary

- The possible values of \( y \) are \( y = 7 \) and \( y = -1 \).
- The radius of the circle is \(\sqrt{793}\).

If you need further details or have any questions, feel free to ask!

### Related Questions

1. How do you derive the equation of a circle from given points?
2. What is the significance of the center of a circle in determining its equation?
3. How can you verify if a point lies on a given circle?
4. What other methods can be used to solve quadratic equations?
5. How is the distance formula derived and what is its importance?
6. Can the distance between points on a circle ever be greater than the diameter?
7. What happens to the circle's radius if the center moves closer to a point on the circle?
8. How would the problem change if the center was a variable point?

**Tip:** Remember to check your work by substituting solutions back into the original equation.

Learn how to find the values of y for points A(-1, y) and B(5, 7) lying on a circle with center O(2, -3y). Includes step-by-step solution using distance formula and quadratic equations. Suitable for high school students studying circle geometry and quadratic equations.

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