From everyday questions to specialized queries, IDNLearn.com has the answers. Discover the information you need quickly and easily with our reliable and thorough Q&A platform.
Sagot :
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]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Your search for answers ends at IDNLearn.com. Thanks for visiting, and we look forward to helping you again soon.
