The diameter of a circle has length 12. The center is at (-5,2) . Give the equation of the circle?
1 answer:
9514 1404 393
Answer:
(x +5)² +(y -2)² = 36
Step-by-step explanation:
The radius is half the diameter, so is 12/2 = 6.
The standard form equation for a circle is ...
(x -h)² +(y -k)² = r² . . . . . . circle centered at (h, k) with radius r
For the given parameters, the equation is ...
(x +5)² +(y -2)² = 36
You might be interested in
>= means greater than or equal to
<= means less than or equal to
---------------------------------------------
Part A
The graph of y >= -3x+3 will have a solid boundary line and the shading will be above the boundary line.
The boundary line y = -3x+3 has a negative slope so it moves down as you read it from left to right. It goes through the points (0,3) and (1,0)
--------------
The graph of y < (3/2)x - 6 will have a dashed or dotted boundary line. The shading is below the boundary.
The graph y = (3/2)x-6 goes through the two points (0,-6) and (2,-3)
--------------
If you graph both y >= -3x+3 and y < (3/2)x - 6 together, you get what you see in the attached image. This solution shaded region is the result of the overlapping prior shaded regions.
---------------------------------------------
Part B
Plug (x,y) = (-6,3) into each inequality to see if we get a true inequality or not
For the first inequality we have
y >= -3x+3
3 >= -3*(-6)+3
3 >= 18+3
3 >= 21
which is false. The value 3 is not larger or equal to 21. So right off the bat we know that (-6,3) is NOT a solution. It is NOT in the solution region.
Let's check the other inequality just for the sake of completeness
y < (3/2)x - 6
3 < (3/2)*(-6) - 6
3 < -9 - 6
3 < -15
this is also false. The value -15 is smaller than 3, since it is to the left of 3
We're given more evidence that (-6,3) is NOT in the solution area. It is outside of both shaded areas.
<= means less than or equal to
---------------------------------------------
Part A
The graph of y >= -3x+3 will have a solid boundary line and the shading will be above the boundary line.
The boundary line y = -3x+3 has a negative slope so it moves down as you read it from left to right. It goes through the points (0,3) and (1,0)
--------------
The graph of y < (3/2)x - 6 will have a dashed or dotted boundary line. The shading is below the boundary.
The graph y = (3/2)x-6 goes through the two points (0,-6) and (2,-3)
--------------
If you graph both y >= -3x+3 and y < (3/2)x - 6 together, you get what you see in the attached image. This solution shaded region is the result of the overlapping prior shaded regions.
---------------------------------------------
Part B
Plug (x,y) = (-6,3) into each inequality to see if we get a true inequality or not
For the first inequality we have
y >= -3x+3
3 >= -3*(-6)+3
3 >= 18+3
3 >= 21
which is false. The value 3 is not larger or equal to 21. So right off the bat we know that (-6,3) is NOT a solution. It is NOT in the solution region.
Let's check the other inequality just for the sake of completeness
y < (3/2)x - 6
3 < (3/2)*(-6) - 6
3 < -9 - 6
3 < -15
this is also false. The value -15 is smaller than 3, since it is to the left of 3
We're given more evidence that (-6,3) is NOT in the solution area. It is outside of both shaded areas.
Answer:
Step-by-step explanation:
The equation of a circle in standard form is
You are given
In order to put the equation in standard from, we need to complete the square. Since there is no y term, the y part is simply y^2, and there is no need to complete the square for y. For x, we do have an x term, so we must complete the square in x.
Start by grouping the x terms and subtracting 45 from both sides.
Now we need to complete the square for x.
The number that completes the square will go in the blank above, and it will also be added to the right side of the equation.
To find the number you need to add to complete the square, take the coefficient of the x term. It is -18. Divide it by 2. You get -9. Now square -9 to get 81. The number that completes the square in x is 81. Now you add it to both sides of the equation.
Answer: