Let p and q be constants such that the graph of x^2+y^2-6x+py+q=0 is tangent to the y-axis. What is the area of the region enclosed by the graph?
I have no idea what to do? Can someone please help me with the solution to the problem?
x^2+y^2-6x+py+q=0
(x-3)^2 + (y+p/2)^2=-q-9-P^2/4
so radius of this circle is
sqrt(q-9-p^2/4)
area= PI (q-9-p^2/4)
check my thinking.
For the area, wouldn't be (q-9-p^2/4)^2 because the formula for the circle is pi radius squared?
To find the area enclosed by the graph, we first need to determine the equation of the circle.
The given equation can be rewritten as follows:
x^2 - 6x + y^2 + py + q = 0
Next, we can complete the square to express the equation in standard form for a circle. We will focus on the terms involving x and y.
x^2 - 6x + 9 + y^2 + py + q = 9
Rearranging the terms:
(x^2 - 6x + 9) + (y^2 + py) = 9 - q
We can now rewrite the quadratic terms by completing the squares:
(x - 3)^2 + (y^2 + py) = 9 - q
To form a perfect square trinomial, we add ((p/2)^2) = (p^2)/4 to both sides:
(x - 3)^2 + (y^2 + py) + (p^2)/4 = 9 - q + (p^2)/4
To make the equation represent a circle centered at (h, k), we need the additional term on the left side. Comparing this with the standard equation for a circle:
(x - h)^2 + (y - k)^2 = r^2
We can conclude that the circle's center is (3, -p/2) and the radius squared is 9 - q + (p^2)/4.
Now, the equation states that the circle is tangent to the y-axis. A circle is tangent to the y-axis when its center lies on the y-axis. Therefore, the x-coordinate of the center, which is 3, must be zero. Hence, p/2 = 0, so p = 0.
Now, let's determine the radius squared:
r^2 = 9 - q + (p^2)/4
r^2 = 9 - q
Since p = 0, r^2 = 9 - q.
Finally, we can calculate the area enclosed by the graph, which is the area of the circle:
Area = πr^2 = π(9 - q)
Therefore, the area of the region enclosed by the graph is π(9 - q).
To find the area enclosed by the graph, we first need to determine the equation of the circle represented by the given equation. Let's walk through the solution step by step:
Step 1: Rewrite the equation:
The given equation can be rearranged to represent the standard form of a circle equation:
x^2 - 6x + y^2 + py + q = 0
Step 2: Complete the square for x terms:
To complete the square for the x terms, we need to add and subtract the square of half of the coefficient of x. In this case, the coefficient of x is -6:
x^2 - 6x + 9 + y^2 + py + q = 9
Step 3: Group the terms:
Rearrange the equation to group the x terms and the y terms separately:
(x^2 - 6x + 9) + (y^2 + py + q) = 9
Step 4: Factor the squared terms:
Factor the quadratic trinomial (x^2 - 6x + 9) as the square of a binomial:
(x - 3)^2 + (y^2 + py + q) = 9
Step 5: Simplify the equation:
Rearrange the equation and collect any remaining constants:
(x - 3)^2 + y^2 + py + (q - 9) = 0
Step 6: Determine the equation of the circle:
We can now deduce the equation of the circle from the standard equation form:
(x - h)^2 + (y - k)^2 = r^2
Comparing the given equation with the standard equation, we can deduce the following values:
Center of the circle (h, k) = (3, 0)
Radius of the circle r^2 = q - 9
Step 7: Find the value of q to ensure the circle is tangent to the y-axis:
Since the graph is tangent to the y-axis, the radius must be equal to the x-coordinate of the center of the circle. Therefore, r^2 = 3^2 = 9.
Setting the values of r^2 equal to each other, we have q - 9 = 9, implying that q = 18.
Step 8: Calculate the area of the region enclosed by the graph:
The area of a circle is given by the formula A = πr^2. Plugging in the value of the radius r = 3 into the formula, we get A = π(3^2) = 9π.
Therefore, the area of the region enclosed by the graph is 9π square units.