Answer :
Let's solve the given problem step-by-step.
### (i) Find the radius and the coordinates of the center of the circle.
Given the equation of the circle is:
[tex]\[ x^2 + y^2 + 6x - 8y + 9 = 0 \][/tex]
First, we rewrite this equation in the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex] by completing the square.
1. Complete the square for the [tex]\(x\)[/tex] terms:
[tex]\[ x^2 + 6x \][/tex]
We add and subtract [tex]\( \left(\frac{6}{2}\right)^2 = 9\)[/tex]:
[tex]\[ (x + 3)^2 - 9 \][/tex]
2. Complete the square for the [tex]\(y\)[/tex] terms:
[tex]\[ y^2 - 8y \][/tex]
We add and subtract [tex]\( \left(\frac{-8}{2}\right)^2 = 16\)[/tex]:
[tex]\[ (y - 4)^2 - 16 \][/tex]
3. Substitute these into the original equation:
[tex]\[ (x + 3)^2 - 9 + (y - 4)^2 - 16 + 9 = 0 \][/tex]
Simplify:
[tex]\[ (x + 3)^2 + (y - 4)^2 - 16 = 0 \][/tex]
[tex]\[ (x + 3)^2 + (y - 4)^2 = 16 \][/tex]
Now, comparing the standard form:
- The center [tex]\((h, k)\)[/tex] of the circle is at [tex]\((-3, 4)\)[/tex].
- The radius [tex]\(r\)[/tex] of the circle is [tex]\(\sqrt{16} = 4\)[/tex].
So, the radius and the coordinates of the center are [tex]\((-3, 4)\)[/tex] and [tex]\(4\)[/tex], respectively.
### (ii) Given that [tex]\(MN\)[/tex] is a diameter of the circle and the coordinates of [tex]\(M\)[/tex] are [tex]\((1, 4)\)[/tex], find the coordinates of [tex]\(N\)[/tex].
The center of the circle is the midpoint of the diameter [tex]\(MN\)[/tex]. Let the coordinates of [tex]\(N\)[/tex] be [tex]\((x, y)\)[/tex].
Using the midpoint formula for the line segment from [tex]\(M(1, 4)\)[/tex] to [tex]\(N(x, y)\)[/tex]:
[tex]\[ \left( \frac{1 + x}{2}, \frac{4 + y}{2} \right) = (-3, 4) \][/tex]
We can now set up and solve the equations:
[tex]\[ \frac{1 + x}{2} = -3 \quad \Rightarrow \quad 1 + x = -6 \quad \Rightarrow \quad x = -7 \][/tex]
[tex]\[ \frac{4 + y}{2} = 4 \quad \Rightarrow \quad 4 + y = 8 \quad \Rightarrow \quad y = 4 \][/tex]
Thus, the coordinates of [tex]\(N\)[/tex] are [tex]\((-7, 4)\)[/tex].
### (iii) Show that the circle [tex]\(C\)[/tex] is tangent to the x-axis.
For the circle to be tangent to the x-axis, the shortest distance from the center of the circle to the x-axis must be equal to the radius of the circle.
The center of the circle is at [tex]\((-3, 4)\)[/tex]. The shortest distance to the x-axis is the absolute value of the y-coordinate of the center, which is [tex]\(4\)[/tex].
Since the radius of the circle is [tex]\(4\)[/tex], it is indeed tangent to the x-axis.
### (iv) The circle is reflected in the line [tex]\(x = 2\)[/tex]. Find the equation that represents the reflected circle.
To find the equation of the circle reflected in the line [tex]\(x = 2\)[/tex], we reflect the center of the circle [tex]\((-3, 4)\)[/tex] across the line [tex]\(x = 2\)[/tex].
The x-coordinate of the reflected center [tex]\((h', k)\)[/tex] is given by:
[tex]\[ h' = 2 \cdot 2 - (-3) = 4 + 3 = 7 \][/tex]
The y-coordinate remains the same, so the reflected center is [tex]\((7, 4)\)[/tex].
The radius remains the same at [tex]\(4\)[/tex].
Thus, the equation of the reflected circle is:
[tex]\[ (x - 7)^2 + (y - 4)^2 = 16 \][/tex]
So, the solution includes:
- Center: [tex]\((-3, 4)\)[/tex]
- Radius: [tex]\(4\)[/tex]
- Coordinates of [tex]\(N\)[/tex]: [tex]\((-7, 4)\)[/tex]
- The circle is tangent to the x-axis [tex]\( (\text{proven}) \)[/tex]
- Equation of reflected circle: [tex]\( (x - 7)^2 + (y - 4)^2 = 16\)[/tex]
### (i) Find the radius and the coordinates of the center of the circle.
Given the equation of the circle is:
[tex]\[ x^2 + y^2 + 6x - 8y + 9 = 0 \][/tex]
First, we rewrite this equation in the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex] by completing the square.
1. Complete the square for the [tex]\(x\)[/tex] terms:
[tex]\[ x^2 + 6x \][/tex]
We add and subtract [tex]\( \left(\frac{6}{2}\right)^2 = 9\)[/tex]:
[tex]\[ (x + 3)^2 - 9 \][/tex]
2. Complete the square for the [tex]\(y\)[/tex] terms:
[tex]\[ y^2 - 8y \][/tex]
We add and subtract [tex]\( \left(\frac{-8}{2}\right)^2 = 16\)[/tex]:
[tex]\[ (y - 4)^2 - 16 \][/tex]
3. Substitute these into the original equation:
[tex]\[ (x + 3)^2 - 9 + (y - 4)^2 - 16 + 9 = 0 \][/tex]
Simplify:
[tex]\[ (x + 3)^2 + (y - 4)^2 - 16 = 0 \][/tex]
[tex]\[ (x + 3)^2 + (y - 4)^2 = 16 \][/tex]
Now, comparing the standard form:
- The center [tex]\((h, k)\)[/tex] of the circle is at [tex]\((-3, 4)\)[/tex].
- The radius [tex]\(r\)[/tex] of the circle is [tex]\(\sqrt{16} = 4\)[/tex].
So, the radius and the coordinates of the center are [tex]\((-3, 4)\)[/tex] and [tex]\(4\)[/tex], respectively.
### (ii) Given that [tex]\(MN\)[/tex] is a diameter of the circle and the coordinates of [tex]\(M\)[/tex] are [tex]\((1, 4)\)[/tex], find the coordinates of [tex]\(N\)[/tex].
The center of the circle is the midpoint of the diameter [tex]\(MN\)[/tex]. Let the coordinates of [tex]\(N\)[/tex] be [tex]\((x, y)\)[/tex].
Using the midpoint formula for the line segment from [tex]\(M(1, 4)\)[/tex] to [tex]\(N(x, y)\)[/tex]:
[tex]\[ \left( \frac{1 + x}{2}, \frac{4 + y}{2} \right) = (-3, 4) \][/tex]
We can now set up and solve the equations:
[tex]\[ \frac{1 + x}{2} = -3 \quad \Rightarrow \quad 1 + x = -6 \quad \Rightarrow \quad x = -7 \][/tex]
[tex]\[ \frac{4 + y}{2} = 4 \quad \Rightarrow \quad 4 + y = 8 \quad \Rightarrow \quad y = 4 \][/tex]
Thus, the coordinates of [tex]\(N\)[/tex] are [tex]\((-7, 4)\)[/tex].
### (iii) Show that the circle [tex]\(C\)[/tex] is tangent to the x-axis.
For the circle to be tangent to the x-axis, the shortest distance from the center of the circle to the x-axis must be equal to the radius of the circle.
The center of the circle is at [tex]\((-3, 4)\)[/tex]. The shortest distance to the x-axis is the absolute value of the y-coordinate of the center, which is [tex]\(4\)[/tex].
Since the radius of the circle is [tex]\(4\)[/tex], it is indeed tangent to the x-axis.
### (iv) The circle is reflected in the line [tex]\(x = 2\)[/tex]. Find the equation that represents the reflected circle.
To find the equation of the circle reflected in the line [tex]\(x = 2\)[/tex], we reflect the center of the circle [tex]\((-3, 4)\)[/tex] across the line [tex]\(x = 2\)[/tex].
The x-coordinate of the reflected center [tex]\((h', k)\)[/tex] is given by:
[tex]\[ h' = 2 \cdot 2 - (-3) = 4 + 3 = 7 \][/tex]
The y-coordinate remains the same, so the reflected center is [tex]\((7, 4)\)[/tex].
The radius remains the same at [tex]\(4\)[/tex].
Thus, the equation of the reflected circle is:
[tex]\[ (x - 7)^2 + (y - 4)^2 = 16 \][/tex]
So, the solution includes:
- Center: [tex]\((-3, 4)\)[/tex]
- Radius: [tex]\(4\)[/tex]
- Coordinates of [tex]\(N\)[/tex]: [tex]\((-7, 4)\)[/tex]
- The circle is tangent to the x-axis [tex]\( (\text{proven}) \)[/tex]
- Equation of reflected circle: [tex]\( (x - 7)^2 + (y - 4)^2 = 16\)[/tex]
