Determine an equation of the circle that is tangent to both axes, has its center in the fourth quadrant and passes through the point (8, -9).
Help please?
Thanks much!
well, you know if it is tangent to both axis, the center is at r,-r
It passes through 8,-9 so the distance from that point to the center is also r.
r^2=(8-r)^2 + (-9+r)^2
= 64-16r+r^2+81-18r+r^2
r^2-34r-145=0
(r-29)(r-5)=0
r=29, or r=5
check both answers.
then put it in standard form.
To determine the equation of the circle, let's follow these steps:
Step 1: Find the coordinates of the center of the circle:
- Since the circle is tangent to both axes, the radius of the circle will be equal to the distance from the center to either of the axes.
- Let's assume the coordinates of the center are (h, k).
- Since the center is in the fourth quadrant, both h and k will be positive.
- Since the circle is tangent to the x-axis, the distance from the center to the x-axis is equal to the radius, which is k.
- Similarly, since the circle is tangent to the y-axis, the distance from the center to the y-axis is equal to the radius, which is h.
- Therefore, the coordinates of the center are (h, k) = (k, h).
Step 2: Use the distance formula to find the distance between the center and the given point (8, -9):
- The distance formula is given by √((x2-x1)^2 + (y2-y1)^2).
- Substituting the values for (x1, y1) = (k, h) and (x2, y2) = (8, -9), the equation becomes √((8-k)^2 + (-9-h)^2).
Step 3: Equate the distance obtained in Step 2 with the radius:
- The distance from the center to the point (8, -9) is equal to the radius, which is k (or h).
- Therefore, √((8-k)^2 + (-9-h)^2) = k.
Step 4: Solve the equation from Step 3 for k (or h):
- Squaring both sides of the equation, we get (8-k)^2 + (-9-h)^2 = k^2.
- Expanding and simplifying, we have:
64 - 16k + k^2 + 81 + 18h + h^2 = k^2.
145 - 16k + 18h + h^2 = 0.
- Rearranging the equation, we obtain 18h - 16k = -145.
So, the equation of the circle that is tangent to both axes, has its center in the fourth quadrant, and passes through the point (8, -9) is 18h - 16k = -145.
To determine the equation of the circle, we can use the standard form of the equation of a circle, which is:
(x - h)^2 + (y - k)^2 = r^2
Where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle.
Given that the circle is tangent to both axes, we can infer that the center of the circle lies on the line y = x since it is perpendicular to both axes. Additionally, since the center is in the fourth quadrant, both the x-coordinate and y-coordinate will be negative.
Let's find the center of the circle:
Since the center lies on the line y = x, we can substitute this equation into the standard form. So, the equation becomes:
(x - h)^2 + (x - k)^2 = r^2
Substituting the x-coordinate (-h) and y-coordinate (-k) gives us:
(-h - h)^2 + (-h - k)^2 = r^2
Simplifying, we get:
4h^2 + 2hk + k^2 = r^2 .....(1)
Next, we know that the circle passes through the point (8, -9). So, we can substitute these coordinates into equation (1):
4(8)^2 + 2(8)(-9) + (-9)^2 = r^2
Simplifying, we get:
256 - 144 - 81 = r^2
31 = r^2
Now, we have the radius (r^2 = 31). To determine the equation of the circle, we need to find the center (h, k). We can do this by solving the following system of equations:
Equation 1: 4h^2 + 2hk + k^2 = 31
Equation 2: h = -k
Substituting Equation 2 into Equation 1 gives us:
4(-k)^2 + 2(-k)k + k^2 = 31
Simplifying, we get:
4k^2 - 2k^2 + k^2 = 31
3k^2 = 31
Dividing both sides by 3, we get:
k^2 = 31/3
Taking the square root of both sides, we get:
k = ±√(31/3)
Since the center lies in the fourth quadrant (negative x-coordinate and negative y-coordinate), we take the negative square root:
k = -√(31/3)
Now, substituting this value of k into Equation 2 to get h, we get:
h = -(-√(31/3)) = √(31/3)
So, the center of the circle is (√(31/3), -√(31/3)).
Finally, we can substitute the values of the center and radius into the equation of the circle to get the final equation:
(x - √(31/3))^2 + (y + √(31/3))^2 = 31
Therefore, the equation of the circle that is tangent to both axes, has its center in the fourth quadrant, and passes through the point (8, -9) is:
(x - √(31/3))^2 + (y + √(31/3))^2 = 31