A line passes through the points with coordinates (8,1) and (10,0).

A circle centred on the origin also passes through the point with coordinates (8,1).

Work out the coordinates of the other point where the line and the circle meet.

Let's solve the problem step by step.

### Step 1: Equation of the line passing through (8,1) and (10,0)
To find the equation of the line, we use the slope-intercept form, \( y = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept.

#### Finding the slope (\( m \)):
The slope \( m \) is given by:
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 1}{10 - 8} = \frac{-1}{2}
\]
So, \( m = -\frac{1}{2} \).

#### Using the point-slope form to find the equation:
Using the point (8,1):
\[
y - y_1 = m(x - x_1)
\]
\[
y - 1 = -\frac{1}{2}(x - 8)
\]
\[
y - 1 = -\frac{1}{2}x + 4
\]
\[
y = -\frac{1}{2}x + 5
\]
So, the equation of the line is:
\[
y = -\frac{1}{2}x + \frac{9}{2}
\]

### Step 2: Equation of the circle centered at the origin and passing through (8,1)
The equation of a circle centered at the origin \((0, 0)\) with radius \(r\) is:
\[
x^2 + y^2 = r^2
\]
Since the circle passes through the point \((8, 1)\), we substitute these coordinates into the equation to find \(r^2\):
\[
8^2 + 1^2 = r^2
\]
\[
64 + 1 = r^2
\]
\[
r^2 = 65
\]
So, the equation of the circle is:
\[
x^2 + y^2 = 65
\]

### Step 3: Find the other intersection point of the line and the circle
To find the other point of intersection, we substitute the equation of the line \( y = -\frac{1}{2}x + \frac{9}{2} \) into the equation of the circle \( x^2 + y^2 = 65 \).

Substitute \( y = -\frac{1}{2}x + \frac{9}{2} \) into \( x^2 + y^2 = 65 \):
\[
x^2 + \left(-\frac{1}{2}x + \frac{9}{2}\right)^2 = 65
\]
Expanding the square:
\[
x^2 + \left(\frac{1}{4}x^2 - \frac{9}{2}x + \frac{81}{4}\right) = 65
\]
Simplifying:
\[
x^2 + \frac{1}{4}x^2 - \frac{9}{2}x + \frac{81}{4} = 65
\]
Multiply through by 4 to eliminate the fraction:
\[
4x^2 + x^2 - 18x + 81 = 260
\]
Simplify:
\[
5x^2 - 18x - 179 = 0
\]
Now, solve this quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Where \( a = 5 \), \( b = -18 \), and \( c = -179 \):
\[
x = \frac{18 \pm \sqrt{(-18)^2 - 4(5)(-179)}}{10}
\]
\[
x = \frac{18 \pm \sqrt{324 + 3580}}{10}
\]
\[
x = \frac{18 \pm \sqrt{3904}}{10}
\]
\[
x = \frac{18 \pm 62.48}{10}
\]
This gives two values for \(x\):
\[
x_1 = \frac{80.48}{10} = 8.048, \quad x_2 = \frac{-44.48}{10} = -4.448
\]
Since \( (8,1) \) is already known, the other \( x \)-coordinate is \( x_2 = -4.448 \).

### Step 4: Find the corresponding \( y \)-coordinate
Substitute \( x = -4.448 \) back into the equation of the line:
\[
y = -\frac{1}{2}(-4.448) + \frac{9}{2}
\]
\[
y = 2.224 + 4.5 = 6.724
\]

So, the coordinates of the other point where the line and the circle meet are approximately:
\[
(-4.448, 6.724)
\]

Would you like more details or have any questions?

Here are five related questions:
1. How do you derive the general form of a line equation from two points?
2. What is the significance of the slope in the line equation?
3. How do you find the radius of a circle given its equation?
4. How does substituting the line equation into the circle equation help find intersection points?
5. Can you solve the quadratic equation using a different method like completing the square?

**Tip:** When solving for intersection points between curves, always ensure to simplify your algebraic expressions step by step to avoid mistakes.

Learn how to find the intersection point of a line passing through (8,1) and (10,0) with a circle centered at the origin and passing through (8,1). Detailed step-by-step solution provided.

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