find the equation of circle having centre in 1st quadrant, touching x-axis,having a common tangent y=3^1/2x+4 with the circle x^2+y^2+4x+4y+4=0 such that the distance between two circles along the x-axis is 3 units ?
To find the equation of the circle, we need to determine its center and radius.
Let's start by analyzing the given information:
1. The circle's center lies in the first quadrant and touches the x-axis. Let's assume the center coordinates as (a, 0).
2. The common tangent of the circle and the line y = √3x + 4 is y = √3x + 4. This tangent line is perpendicular to the radius of the circle at the point of tangency.
3. The given circle has the equation x^2 + y^2 + 4x + 4y + 4 = 0.
Now, let's proceed step-by-step:
Step 1: Determine the center of the circle.
Since the circle lies in the first quadrant and touches the x-axis, we can conclude that the center's y-coordinate is 0. Therefore, the center of the circle is (a, 0).
Step 2: Find the point of tangency.
To find the point of tangency between the given line and the circle, we need to solve the system of equations formed by equating the equation of the line and the equation of the circle. Let's substitute y = √3x + 4 into the circle equation:
x^2 + (√3x + 4)^2 + 4x + 4(√3x + 4) + 4 = 0
Simplifying the equation:
x^2 + 3x^2 + 8√3x + 16 + 4x + 4√3x + 16 + 4 = 0
Combining like terms:
4x^2 + (12 + 12√3)x + 40 = 0
Step 3: Use the discriminant to determine if there is a unique solution.
To determine if the quadratic equation has real solutions, we use the discriminant (b^2 - 4ac) to evaluate whether it is positive, zero, or negative.
The discriminant D = (12 + 12√3)^2 - 4*4*40 simplifies to 144 + 288√3 + 432 - 640 = 400 + 288√3.
Since the discriminant is positive, there are two distinct real solutions for x.
Step 4: Use the solutions for x to find the y-coordinates of the point of tangency.
Substitute the obtained solutions for x back into the equation of the line y = √3x + 4 to find the y-values of the point(s) of tangency.
For the point of tangency (x₁, y₁):
y₁ = √3x₁ + 4
For the point of tangency (x₂, y₂):
y₂ = √3x₂ + 4
Step 5: Determine the radius of the circle.
Since the distance between the two circles along the x-axis is 3 units, the difference between the x-coordinates of the point(s) of tangency and the x-coordinate of the circle's center is ±3 units.
Let's consider the point of tangency with the smaller x-coordinate:
x₁ - a = 3
Solving for a:
a = x₁ - 3
Step 6: Form the equation of the circle.
The general equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius.
Using the center coordinates (a, 0) and the point of tangency (x₁, y₁), we have:
(x - a)^2 + (y - 0)^2 = (x - a)^2 + y^2
Since the center is at (a, 0), we have:
(x - a)^2 + y^2 = r^2
The radius r can be found using the distance formula between the center (a, 0) and the point of tangency (x₁, y₁):
r = √((x₁ - a)^2 + y₁^2)
So, the equation of the circle is:
(x - a)^2 + y^2 = ((x₁ - a)^2 + y₁^2)
You can substitute the values of x₁, y₁, and a into the equation to obtain the final equation of the circle.
To find the equation of the circle, we need to follow a few steps:
Step 1: Find the center of the circle
Given that the circle is in the first quadrant and touches the x-axis, we can conclude that the center lies on the positive y-axis. Let's assume the coordinates of the center are (0, a).
Step 2: Find the radius of the circle
Since the given circle x^2+y^2+4x+4y+4=0 touches the line y=3^1/2x+4, we can equate the distances from the center of the circle to these two lines. The distance from the center to the line y=3^1/2x+4 is given as:
Distance = (|a - 3^1/2 * 0 - 4|) / sqrt(1 + 3^1/2^2)
= (|a - 4|) / sqrt(1 + 3)
= (|a - 4|) / sqrt(4)
= (|a - 4|) / 2
The radius of the given circle is sqrt(4) = 2. Since the distance between the two circles along the x-axis is 3 units, the radius of the other circle is 2 + 3 = 5.
Step 3: Use the center and radius to form the equation of the circle
The equation of a circle with center (h, k) and radius r is given as:
(x - h)^2 + (y - k)^2 = r^2
Using the center (0, a) and radius 5, we get:
(x - 0)^2 + (y - a)^2 = 5^2
x^2 + (y - a)^2 = 25
Thus, the equation of the circle is x^2 + (y - a)^2 = 25, where a is the y-coordinate of the circle's center in the first quadrant. It remains to determine the specific value of 'a'.
To find the value of 'a', substitute the equation of the circle into the equation of the given circle x^2 + y^2 + 4x + 4y + 4 = 0 and solve for 'a'.
By substituting x^2 + (y - a)^2 = 25 into x^2 + y^2 + 4x + 4y + 4 = 0, we get:
x^2 + y^2 + 4x + 4y + 4 = x^2 + (y - a)^2
4x + 4y + 4 = (y - a)^2
Expanding and simplifying further, we get:
4x + 4y + 4 = y^2 - 2ay + a^2
Rearranging the equation, we have:
(y^2 - 2ay) + (4y - 4) - (4x) + (a^2 - 4) = 0
For the equation to represent a tangent, the discriminant of y must be zero:
(-2a)^2 - 4(1)(4y - 4)(a^2 - 4) = 0
Simplifying the equation, we get:
4a^2 - 4(4y - 4)(a^2 - 4) = 0
Expand the equation further:
4a^2 - 16(a^2 - 4y + 4) = 0
4a^2 - 16a^2 + 64y - 64 = 0
-12a^2 + 64y - 64 = 0
Since the equation implies (a, y) are the coordinates of the point of tangency, this point lies on the line y = 3^1/2x + 4. Substitute this line equation into the equation above:
-12a^2 + 64(3^1/2a + 4) - 64 = 0
Simplify the equation:
-12a^2 + 192a + 160 - 64 = 0
-12a^2 + 192a + 96 = 0
-3a^2 + 48a + 24 = 0
Solve this quadratic equation to find the value(s) of 'a'.
Is that y = (√3)x+4 ?
The other circle is clearly
(x+2)^2 + (y+2)^2 = 4
and the way the question is worded, it appears that the first circle is also tangent to the x-axis at (1,0), making it
(x-1)^2 + (y-k)^2 = k^2
Now it should not be too hard to find k so that the desired line is a common tangent.