Find the equation of a circle with radius 4 units ,whose center lies on the line 13x+4y=32 and which touch the line 3x+4y+28= 0.
One way is to find the point on 13x+4y=32 which is at a distance of 4 from the line 3x+4y=28.
Since the distance from (h,k) to ax+by+c=0 is
|ah+bk+c|/√(a^2+b^2)
and we know that k = (32-13h)/4, we have
|3h+32-13h+28|/5 = 4
|-10h+60| = 20
h = 4 or 8
so, k = -5 or -18
Check the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3D%2832-13x%29%2F4%2Cy%3D%28-28-3x%29%2F4%2C%28x-4%29^2%2B%28y%2B5%29^2%3D16%2C%28x-8%29^2%2B%28y%2B18%29^2%3D16
To find the equation of the circle, we need to determine the coordinates of its center first.
We know that the center of the circle lies on the line 13x + 4y = 32.
To find the coordinates, let's rewrite the equation of the line in slope-intercept form:
4y = -13x + 32
Dividing both sides by 4, we get:
y = -13/4x + 8
We can identify the slope of the line as -13/4,
So, any line perpendicular to this will have a slope of 4/13.
Now, let's find the equation of the line perpendicular to 13x + 4y = 32 and passes through its midpoint.
To find the midpoint, we need to find two points on the line 13x + 4y = 32.
Let's set x = 0 to find the y-coordinate of the midpoint:
13(0) + 4y = 32
4y = 32
y = 8
So, one point on the line is (0, 8).
Now, let's set x = 1 to find another point on the line:
13(1) + 4y = 32
13 + 4y = 32
4y = 32 - 13
4y = 19
y = 19/4
So, the second point on the line is (1, 19/4).
The midpoint of these two points is ((0 + 1)/2, (8 + 19/4)/2)
= (1/2, 35/8)
Now, we have the center of the circle.
Since we know that the radius of the circle is 4 units, and it touches the line 3x + 4y + 28 = 0, we can find the equation of the circle using the distance formula.
The distance between the center of the circle and the line is equal to the radius.
The equation of the line is in the form ax + by + c = 0, where a = 3, b = 4, and c = 28.
The formula for the distance between a point (x₁, y₁) and a line ax + by + c = 0 is:
d = |ax₁ + by₁ + c| / sqrt(a² + b²)
Plugging in the values, we get:
4 = |3(1/2) + 4(35/8) + 28| / sqrt(3² + 4²)
4 = |-15/2 + 140/8 + 28| / sqrt(9 + 16)
4 = |(-30 + 140 + 224)/8| / sqrt(25)
4 = |334/8| / 5
4 = 334/40
Now, let's find the equation of the circle given its center (h, k) = (1/2, 35/8) and the radius r = 4:
The equation of a circle is given by:
(x - h)² + (y - k)² = r²
Substituting the values, we get:
(x - 1/2)² + (y - 35/8)² = 4²
(x - 1/2)² + (y - 35/8)² = 16
Therefore, the equation of the circle is:
(x - 1/2)² + (y - 35/8)² = 16
To find the equation of a circle, we need the coordinates of its center and the radius.
First, let's find the center of the circle. Since the center lies on the line 13x + 4y = 32, we can express y in terms of x using this equation.
Rearrange the equation 13x + 4y = 32 to solve for y:
4y = -13x + 32
y = (-13/4)x + 8
Now we have an equation in slope-intercept form (y = mx + b), where the coefficient of x is the slope and b is the y-intercept.
Since the center of the circle lies on this line, we only have one unknown coordinate, which can be represented as (x, (-13/4)x + 8). Let's call this point (h, k), where h is the x-coordinate and k is the y-coordinate of the center.
Next, we need to determine the radius of the circle. The radius is given as 4 units.
Now we have all the information we need to write the equation of the circle. The general equation of a circle with center (h, k) and radius r is:
(x - h)^2 + (y - k)^2 = r^2
Substituting the values we have:
(x - h)^2 + (y - k)^2 = 4^2
Now, let's find the point where the circle touches the line 3x + 4y + 28 = 0. To find this point, we need to solve the simultaneous equations formed by the equation of the circle and the equation of the line.
Substitute the expression for y from the circle equation into the line equation:
3x + 4((-13/4)x + 8) + 28 = 0
Simplify:
3x - 13x + 32 + 28 = 0
Combine like terms:
-10x + 60 = 0
Solve for x:
-10x = -60
x = 6
Substitute this value back into the expression for y:
y = (-13/4)(6) + 8
y = -19/2 + 8
y = -19/2 + 16/2
y = -3/2
So, the point of contact between the circle and the line is (6, -3/2).
Now, we have the center (h, k) = (6, -3/2) and the radius r = 4 units.
Substituting these values into the equation of the circle:
(x - 6)^2 + (y + 3/2)^2 = 4^2
Expanding and simplifying:
(x - 6)^2 + (y + 3/2)^2 = 16
This is the equation of the circle with radius 4 units, whose center lies on the line 13x + 4y = 32 and which touches the line 3x + 4y + 28 = 0.